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### Topics You Need To Understand For This Page

 limits infinite limits

### Related Topics and Links

 sequences youtube playlist Math Is Fun - Geometric Sequences and Sums Pauls Online Math Notes - Sequences

### 17Calculus Subjects Listed Alphabetically

Single Variable Calculus

 Absolute Convergence Alternating Series Arc Length Area Under Curves Chain Rule Concavity Conics Conics in Polar Form Conditional Convergence Continuity & Discontinuities Convolution, Laplace Transforms Cosine/Sine Integration Critical Points Cylinder-Shell Method - Volume Integrals Definite Integrals Derivatives Differentials Direct Comparison Test Divergence (nth-Term) Test
 Ellipses (Rectangular Conics) Epsilon-Delta Limit Definition Exponential Derivatives Exponential Growth/Decay Finite Limits First Derivative First Derivative Test Formal Limit Definition Fourier Series Geometric Series Graphing Higher Order Derivatives Hyperbolas (Rectangular Conics) Hyperbolic Derivatives
 Implicit Differentiation Improper Integrals Indeterminate Forms Infinite Limits Infinite Series Infinite Series Table Infinite Series Study Techniques Infinite Series, Choosing a Test Infinite Series Exam Preparation Infinite Series Exam A Inflection Points Initial Value Problems, Laplace Transforms Integral Test Integrals Integration by Partial Fractions Integration By Parts Integration By Substitution Intermediate Value Theorem Interval of Convergence Inverse Function Derivatives Inverse Hyperbolic Derivatives Inverse Trig Derivatives
 Laplace Transforms L'Hôpital's Rule Limit Comparison Test Limits Linear Motion Logarithm Derivatives Logarithmic Differentiation Moments, Center of Mass Mean Value Theorem Normal Lines One-Sided Limits Optimization
 p-Series Parabolas (Rectangular Conics) Parabolas (Polar Conics) Parametric Equations Parametric Curves Parametric Surfaces Pinching Theorem Polar Coordinates Plane Regions, Describing Power Rule Power Series Product Rule
 Quotient Rule Radius of Convergence Ratio Test Related Rates Related Rates Areas Related Rates Distances Related Rates Volumes Remainder & Error Bounds Root Test Secant/Tangent Integration Second Derivative Second Derivative Test Shifting Theorems Sine/Cosine Integration Slope and Tangent Lines Square Wave Surface Area
 Tangent/Secant Integration Taylor/Maclaurin Series Telescoping Series Trig Derivatives Trig Integration Trig Limits Trig Substitution Unit Step Function Unit Impulse Function Volume Integrals Washer-Disc Method - Volume Integrals Work

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 Acceleration Vector Arc Length (Vector Functions) Arc Length Function Arc Length Parameter Conservative Vector Fields Cross Product Curl Curvature Cylindrical Coordinates
 Directional Derivatives Divergence (Vector Fields) Divergence Theorem Dot Product Double Integrals - Area & Volume Double Integrals - Polar Coordinates Double Integrals - Rectangular Gradients Green's Theorem
 Lagrange Multipliers Line Integrals Partial Derivatives Partial Integrals Path Integrals Potential Functions Principal Unit Normal Vector
 Spherical Coordinates Stokes' Theorem Surface Integrals Tangent Planes Triple Integrals - Cylindrical Triple Integrals - Rectangular Triple Integrals - Spherical
 Unit Tangent Vector Unit Vectors Vector Fields Vectors Vector Functions Vector Functions Equations

Differential Equations

 Boundary Value Problems Bernoulli Equation Cauchy-Euler Equation Chebyshev's Equation Chemical Concentration Classify Differential Equations Differential Equations Euler's Method Exact Equations Existence and Uniqueness Exponential Growth/Decay
 First Order, Linear Fluids, Mixing Fourier Series Inhomogeneous ODE's Integrating Factors, Exact Integrating Factors, Linear Laplace Transforms, Solve Initial Value Problems Linear, First Order Linear, Second Order Linear Systems
 Partial Differential Equations Polynomial Coefficients Population Dynamics Projectile Motion Reduction of Order Resonance
 Second Order, Linear Separation of Variables Slope Fields Stability Substitution Undetermined Coefficients Variation of Parameters Vibration Wronskian

### Search Practice Problems

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17calculus > infinite series > sequences

 Understanding Sequences Notation Arithmetic Sequence Geometric Sequence Sequence Convergence and Divergence Using L'Hôpital's Rule Binomial Theorem Practice

Infinite sequences are an important introduction to series, since infinite series are built upon the concept of sequences.

Sequence Series A sequence is just a list of items separated by commas. With a series, we actually add up some (or all) terms of some sequence.

Sequences can be infinite (have an infinite number of terms) or finite (have a finite number of terms).

 Example 1: $$\{ 1, 2, 3, 4, 5 \}$$ This is a finite sequence since it has a finite number of elements, i.e. after the number $$5$$ there are no more numbers in the sequence and it stops. Example 2: $$\{ 1, 4, 9, 16, 25, 36, . . . \}$$ This is an infinite sequence since it has an infinite number of elements. The infinite part is denoted by the three periods at the end of the list.

We don't always list the elements of a sequence. If there is a logical pattern, we can write it more compact form as an equation. This idea is discussed in the next section on notation.

Notation

Sequences may be written in several different ways. These are the ones you will come across the most in calculus.
1. As a list of items; we showed this in the two examples above.
2. $$\displaystyle{\{a_n\}_{n=1}^{n=\infty}}$$ or more simply as $$\displaystyle{\{a_n\}_{1}^{\infty}}$$; in example 2 above, we would also include what $$a_n$$ was, i.e. $$a_n = n^2$$.
3. Similar to the last way, we could write the sequence in example 2 above as $$\displaystyle{\{n^2\}_{n=1}^{n=\infty}}$$; in this case, $$a_n$$ is implied to be $$a_n = n^2$$.
4. You also often see a sequence written without specifying $$n$$, i.e $$\{a_n\}$$ and n is implied as $$n = 1, 2, 3, . . . \infty$$
5. A sequence can also be built from itself, called a recursive sequence. In these types of sequences, the first few terms are given and the following terms are defined based on those first terms. For example, $$a_1 = 1, ~~ a_{n+1} = 2a_n$$.
6. Finally, you can describe a sequence with words. This is shown in practice problem 239 below. This is a very unusual way to do it, so you won't see it much.

The whole idea is to emphasize that a sequence is a LIST of elements. Each element stands alone and does not interact at all with any other element.

Here is quick introduction video clip that will get you started. It has a couple of good examples.

### PatrickJMT - Sequences [6min-10secs]

video by PatrickJMT

Arithmetic Sequence

One special type of sequence is an arithmetic sequence, which looks like $$\{ a_n \}$$ where $$a_n = c+nk$$, $$c$$ and $$k$$ are constants. An example is $$\{ 3, 12, 21, 30, . . . \}$$. Notice that in this example, we start with $$3$$ and add $$9$$ to get the next element.

Geometric Sequence

Another special type of sequence is a geometric sequence. This sequence looks like $$\{ a_n \} = \{ c, cr, cr^2, cr^3, . . . \}$$ where the terms are of the form $$a_n = cr^n$$ for $$n=0, 1, 2 . . .$$
$$c$$ is a constant
$$r$$ is a term that does not contain $$n$$.
An example is $$\{ 1, 3, 9, 27, 81, . . . \}$$ where we start with $$1$$ and multiply by $$3$$ to get the next element. Compare this to the arithmetic sequence where we add to get the next element.

Sequence Convergence and Divergence

Convergence of a sequence is defined as the limit of the terms approaching a specific real number. In equation terms, it looks like this. For a sequence $$\{a_n\}$$ where $$n=\{1, 2, 3, 4, . . . \}$$, we calculate $$\displaystyle{ \lim_{n \to \infty}{a_n} = L }$$

Convergence: If $$L$$ is a real number, then we say that the sequence $$\{a_n\}$$ converges to $$L$$.

Divergence: If $$L = \pm \infty$$ or the limit cannot be determined, i.e. it is undefined, then the sequence is said to diverge.

Note: The index $$n$$ can start at zero instead of one. This is what happened with the geometric sequence shown in the previous section. In actuality, the index can start at any number. However, for consistency, we usually start it at either zero or one.

Using L'Hôpital's Rule

When evaluating the limit $$\displaystyle{ \lim_{n \to \infty}{a_n} = L }$$, we sometimes need to use L'Hôpital's Rule. In the course of using L'Hôpital's Rule, we will need to take some derivatives. However, derivatives are only defined on continuous functions. And since a sequence is a set of discontinuous values, we can't directly use L'Hôpital's Rule. Fortunately, we have a theorem that will help us.

Theorem

If we can find a continuous function, call it $$g(x)$$ where $$g(n) = a_n$$ for all $$n$$, then $$\displaystyle{ \lim_{x \to \infty}{g(x)} = \lim_{n \to \infty}{a_n} }$$.

Finding a function $$g(x)$$ is not usually that hard, if $$a_n$$ is given in equation terms. Just replace $$n$$ with $$x$$.

I know this seems kind of picky but you need to get used to following the exact requirements of theorem. That is one reason we suggest that you read through and try to understand proofs in calculus. Okay, let's get down to some details and how to work with sequences. These next two videos will help you a lot.

This video is a bit long but well worth taking the time to watch. It contains a lot of detail about sequences and how to work with them and has plenty of good examples.

### Dr Chris Tisdell - Sequences (part 1) [44min-52secs]

video by Dr Chris Tisdell

This video is a continuation of the previous video and is not as long but it discusses the core of sequences, boundedness and convergence behaviour. You need to watch the previous video first to get the context before watching this one.

### Dr Chris Tisdell - Sequences (part 2) [27min-27secs]

video by Dr Chris Tisdell

Binomial Theorem

The binomial theorem is a way to expand a term in the form $$(a+b)^n$$ into a series. We do not cover this topic extensively at this time but here is a video to get you started with an explanation of the equation and an example.

### PatrickJMT - The Binomial Theorem [11min-20secs]

video by PatrickJMT

### Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

Instructions - - Unless otherwise instructed, write the first few terms of these sequences and determine if they converge or diverge. If not specified, start the index at n=1.

Basic Problems

$$\displaystyle{ \left\{ \frac{2n}{6n-5} \right\} }$$

Problem Statement

Determine convergence or divergence of the sequence $$\displaystyle{ \left\{ \frac{2n}{6n-5} \right\} }$$

Since the limit $$\displaystyle{ \lim_{n \to \infty}{a_n} }$$ is finite, the sequence $$\displaystyle{ a_n = \frac{2n}{6n-5} }$$ converges.

Problem Statement

Determine convergence or divergence of the sequence $$\displaystyle{ \left\{ \frac{2n}{6n-5} \right\} }$$

Solution

To determine if the sequence converges, we evaluate the limit $$\displaystyle{\lim_{n \rightarrow \infty}{a_n} }$$. If this limit is finite, then the sequence converges, otherwise it diverges.
To calculate the limit, we can use either L'Hôpital's Rule
$$\displaystyle{ \lim_{n \rightarrow \infty}{ \frac{2n}{6n-5} } = \lim_{n \rightarrow \infty}{ \frac{2}{6} } = 1/3 }$$
Or, since the numerator and denominator are both polynomials, we can multiply the numerator and denominator by $$1/n$$.
$$\displaystyle{ \lim_{n \rightarrow \infty}{ \frac{2n}{6n-5} \cdot \frac{1/n}{1/n} } = \lim_{n \rightarrow \infty}{ \frac{2}{6-5/n} } }$$
Since we know that $$\displaystyle{ \lim_{n \rightarrow \infty}{5/n} = 0 }$$ then $$\displaystyle{ \lim_{n \rightarrow \infty}{ \frac{2}{6-5/n}} = 2/6 = 1/3 }$$

Since the limit $$\displaystyle{ \lim_{n \to \infty}{a_n} }$$ is finite, the sequence $$\displaystyle{ a_n = \frac{2n}{6n-5} }$$ converges.

$$\displaystyle{ \left\{ \frac{n+1}{n^2} \right\}_{n=1}^{n=\infty} }$$

Problem Statement

$$\displaystyle{ \left\{ \frac{n+1}{n^2} \right\}_{n=1}^{n=\infty} }$$

The sequence can be written
$$\displaystyle{ \left\{ \frac{n+1}{n^2} \right\}_{n=1}^{n=\infty} = \left\{ 2, \frac{3}{4}, \frac{4}{9}, \frac{5}{16}, \ldots \right\} }$$ and the sequence converges to zero.

Problem Statement

$$\displaystyle{ \left\{ \frac{n+1}{n^2} \right\}_{n=1}^{n=\infty} }$$

Solution

First, we will build a table to see if we can get an idea of convergence/divergence. This will also give us the first few terms of the sequence.

 n $$a_n$$ 1 $$\displaystyle{ \frac{1+1}{1^2} = \frac{2}{1} = 2 }$$ 2 $$\displaystyle{ \frac{2+1}{2^2} = \frac{3}{4} }$$ 3 $$\displaystyle{ \frac{3+1}{3^2} = \frac{4}{9} }$$ 4 $$\displaystyle{ \frac{4+1}{4^2} = \frac{5}{16} }$$

Okay, so it looks like the sequence may be converging to zero, but let's calculate the limit to make sure.
The limit we need to calculate is $$\displaystyle{ \lim_{n \to \infty}{\frac{n+1}{n^2}} }$$ and we can either use algebra or L'Hôpital's Rule. Let's use algebra.

 $$\displaystyle{ \lim_{n \to \infty}{\frac{n+1}{n^2}} }$$ $$\displaystyle{\lim_{n \to \infty}{\frac{n+1}{n^2} \frac{1/n^2}{1/n^2}} }$$ $$\displaystyle{\lim_{n \to \infty}{\frac{1/n+1/n^2}{1}} = \frac{0+0}{1} = 0 }$$

The sequence can be written
$$\displaystyle{ \left\{ \frac{n+1}{n^2} \right\}_{n=1}^{n=\infty} = \left\{ 2, \frac{3}{4}, \frac{4}{9}, \frac{5}{16}, \ldots \right\} }$$ and the sequence converges to zero.

$$\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }$$

Problem Statement

$$\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }$$

The sequence $$\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }$$ converges.

Problem Statement

$$\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }$$

Solution

In order to determine convergence or divergence, we need to evaluate the limit $$\displaystyle{ \lim_{n \to \infty}{ \frac{3n^2-1}{10n+4n^2} } }$$. If it converges to a real number, then the sequence also converges. Otherwise it diverges.

 $$\displaystyle{ \lim_{n \to \infty}{ \frac{3n^2-1}{10n+4n^2} } }$$ $$\displaystyle{ \lim_{n \to \infty}{ \frac{3n^2-1}{10n+4n^2} \frac{1/n^2}{1/n^2} } }$$ $$\displaystyle{ \lim_{n \to \infty}{ \frac{3-1/n^2}{10/n+4} } = \frac{3}{4} }$$

Since the limit is a real number, the sequence converges.
Notes:
- In this problem, we divided the numerator and denominator by term that matches the highest term. In this case, the highest term was $$n^2$$. This technique works for polynomials and it doesn't matter if the highest term is in the numerator or the denominator.
- We could have used L'Hôpital's Rule to determine this limit instead, since direct substitution yields $$\infty / \infty$$ which is indeterminate.

The sequence $$\displaystyle{ \left\{ \frac{3n^2-1}{10n+4n^2} \right\}_{n=2}^{n=\infty} }$$ converges.

Find the limit of the sequence $$\{a_n\}$$ where $$\displaystyle{ a_n = \frac{2n}{5n-3} }$$.

Problem Statement

Find the limit of the sequence $$\{a_n\}$$ where $$\displaystyle{ a_n = \frac{2n}{5n-3} }$$.

Solution

### 1321 solution video

video by Krista King Math

Find the limit of the sequence $$\{a_n\}$$ where $$\displaystyle{ a_n = \frac{n^2-n+7}{2n^3+n^2} }$$.

Problem Statement

Find the limit of the sequence $$\{a_n\}$$ where $$\displaystyle{ a_n = \frac{n^2-n+7}{2n^3+n^2} }$$.

Solution

### 1322 solution video

video by Krista King Math

Find the nth term of the sequence $$\{ 1, 4, 9, 16 \ldots \}$$

Problem Statement

Find the nth term of the sequence $$\{ 1, 4, 9, 16 \ldots \}$$

Solution

### 1323 solution video

video by Krista King Math

Find a formula for the general term $$a_n$$ of the sequence $$\displaystyle{ \left\{ \frac{1}{2}, -\frac{4}{3}, \frac{9}{4}, -\frac{16}{5}, \frac{25}{6}, \ldots \right\} }$$

Problem Statement

Find a formula for the general term $$a_n$$ of the sequence $$\displaystyle{ \left\{ \frac{1}{2}, -\frac{4}{3}, \frac{9}{4}, -\frac{16}{5}, \frac{25}{6}, \ldots \right\} }$$

Solution

### 1404 solution video

video by Krista King Math

$$\{ n(n-1) \}$$

Problem Statement

$$\{ n(n-1) \}$$

Solution

### 1328 solution video

video by PatrickJMT

$$\displaystyle{ \left\{ 1 + \left( -\frac{1}{2} \right)^n \right\} }$$

Problem Statement

$$\displaystyle{ \left\{ 1 + \left( -\frac{1}{2} \right)^n \right\} }$$

Solution

### 1215 solution video

video by Krista King Math

Compute $$\displaystyle{ \lim_{n\to\infty}{ \frac{\ln(n)}{n} } }$$

Problem Statement

Compute $$\displaystyle{ \lim_{n\to\infty}{ \frac{\ln(n)}{n} } }$$

Solution

### 1865 solution video

video by Dr Chris Tisdell

Compute $$\displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{n} } }$$ using the result from the previous problem.

Problem Statement

Compute $$\displaystyle{ \lim_{n\to\infty}{ \sqrt[n]{n} } }$$ using the result from the previous problem.

Solution

### 1866 solution video

video by Dr Chris Tisdell

Intermediate Problems

$$\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n} \right\}_{n=0}^{n=\infty} }$$

Problem Statement

$$\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n} \right\}_{n=0}^{n=\infty} }$$

The sequence can be written $$\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n}\right\}_{n=0}^{n=\infty} = \left\{ -1, \frac{1}{3}, \frac{-1}{9}, \frac{1}{27}, \ldots \right\} }$$ and the sequence converges to zero.

Problem Statement

$$\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n} \right\}_{n=0}^{n=\infty} }$$

Solution

First, we will build a table to see if we can get an idea of convergence/divergence. This will also give us the first few terms of the sequence.

 n $$a_n$$ 0 $$\displaystyle{ \frac{(-1)^1}{3^0} = \frac{-1}{1} = -1 }$$ 1 $$\displaystyle{ \frac{(-1)^2}{3^1} = \frac{1}{3} }$$ 2 $$\displaystyle{ \frac{(-1)^3}{3^2} = \frac{-1}{9} }$$ 3 $$\displaystyle{ \frac{(-1)^4}{3^3} = \frac{1}{27} }$$

To determine convergence or divergence, we need to calculate the limit $$\displaystyle{ \lim_{n \to \infty}{\frac{(-1)^n}{3^n}} }$$

As $$n$$ increases, the denominator increases also, so we know that $$\displaystyle{ \lim_{n \to \infty}{\frac{1}{3^n}} = 0 }$$
Looking at the numerator, the values are either 1 or -1. Since the numerator is never zero, we end up with the limit being zero ( since any number times zero is zero ).
Note: As mentioned above the numerator is either 1 or -1 and, we can see from the first few terms, that the sign alternates between consecutive terms. This is called an alternating sequences, referring to this alternating sign. This will be important later when you study alternating series.

The sequence can be written $$\displaystyle{ \left\{ \frac{(-1)^{n+1}}{3^n}\right\}_{n=0}^{n=\infty} = \left\{ -1, \frac{1}{3}, \frac{-1}{9}, \frac{1}{27}, \ldots \right\} }$$ and the sequence converges to zero.

For the sequence $$\displaystyle{\left\{\frac{1}{2n+3}\right\}}$$ determine whether it is increasing, decreasing, monotonic and if it is bounded. Explain your reasoning.

Problem Statement

For the sequence $$\displaystyle{\left\{\frac{1}{2n+3}\right\}}$$ determine whether it is increasing, decreasing, monotonic and if it is bounded. Explain your reasoning.

Solution

When she is trying to prove that the series is decreasing, she uses a circular argument and never really proves it. She starts out well by writing $$\displaystyle{ a_{n+1} \leq a_n }$$ as $$\displaystyle{ \frac{1}{2n+5} \leq \frac{1}{2n+3} }$$ and then she just says this is true. It is better to show this definitively like this.
$$\begin{array}{rcl} \displaystyle{ \frac{1}{2n+5} } & \leq & \displaystyle{ \frac{1}{2n+3} } \\ 2n+3 & \leq & 2n+5 \\ 3 & \leq & 5 \end{array}$$
Since $$3 \leq 5$$ is always true then $$\displaystyle{ a_{n+1} \leq a_n }$$ is also always true. This is the correct way to show the inequality holds.
Again when she discusses boundedness, she doesn't show that the inequalities hold for all n. To do this, you need to set up the equalities and manipulate them like we did above.

### 1320 solution video

video by Krista King Math

$$\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }$$

Problem Statement

$$\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }$$

The sequence $$\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }$$ diverges.

Problem Statement

$$\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }$$

Solution

In order to determine convergence or divergence, we need to evaluate the limit $$\displaystyle{ \lim_{n \to \infty}{ \frac{e^{3n}}{n^2} } }$$. Direct substitution yields $$\infty / \infty$$. Since the numerator is not a polynomial, the best way to evaluate this limit is to use L'Hôpital's Rule.
Now, strictly, we can't use L'Hôpital's Rule directly on limit $$\displaystyle{ \lim_{n \to \infty}{ \frac{e^{3n}}{n^2} } }$$. L'Hôpital's Rule works only on continuous functions since derivatives are involved. But we can invoke the theorem that says $$\displaystyle{ \lim_{n \to \infty}{a_n} = \lim_{x \to \infty}{g(x)} }$$ when we have a function $$g(n) = a_n$$ for all $$n$$.
So, let's define $$\displaystyle{ g(x) = \frac{e^{3x}}{x^2} }$$

 $$\displaystyle{ \lim_{n \to \infty}{ \frac{e^{3n}}{n^2} } }$$ $$\displaystyle{ \lim_{x \to \infty}{ \frac{e^{3x}}{x^2} } }$$ $$\displaystyle{ \lim_{x \to \infty}{ \frac{3e^{3x}}{2x} } }$$ $$\displaystyle{\lim_{x \to \infty}{ \frac{9e^{3x}}{2} } = \infty }$$

In the above work, we had to apply L'Hôpital's Rule twice. We didn't show the work but we tried direct substitution after the first application and found that we still had an indeterminate form. This then allowed us to apply L'Hôpital's Rule a second time.

So we found that the limit was $$\infty$$ and, since this is not a finite real number, the sequence is said to diverge.

The sequence $$\displaystyle{ \left\{ \frac{e^{3n}}{n^2} \right\} }$$ diverges.

$$\displaystyle{ \left\{ \frac{2^n}{3^{n+1}} \right\} }$$

Problem Statement

$$\displaystyle{ \left\{ \frac{2^n}{3^{n+1}} \right\} }$$

Solution

### 1329 solution video

video by PatrickJMT

Write down the first few terms of the sequence $$\displaystyle{ \{a_n\} }$$ where $$a_n$$ is the nth digit of $$e$$ and discuss the convergence or divergence of the sequence.

Problem Statement

Write down the first few terms of the sequence $$\displaystyle{ \{a_n\} }$$ where $$a_n$$ is the nth digit of $$e$$ and discuss the convergence or divergence of the sequence.

The sequence $$\{a_n\}=\{ 2, 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, 9, \ldots \}$$ where $$a_n$$ is the nth digit of the irrational number $$e$$. This sequence diverges.

Problem Statement

Write down the first few terms of the sequence $$\displaystyle{ \{a_n\} }$$ where $$a_n$$ is the nth digit of $$e$$ and discuss the convergence or divergence of the sequence.

Solution

We know from algebra that the irrational number $$e = 2.718281828459 \ldots$$. So the sequence is $$\{ 2, 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, 9, \ldots \}$$.
Since $$e$$ is irrational, it will never stop or repeat. Therefore the sequence diverges.

The sequence $$\{a_n\}=\{ 2, 7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, 9, \ldots \}$$ where $$a_n$$ is the nth digit of the irrational number $$e$$. This sequence diverges.