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effective study techniques

In our experience, the Root Test is the least used series test to test for convergence or divergence (which is why it appears last in the infinite series table). The reason is that it is used only in very specific cases, whereas the other tests can be used for a broader range of problems. However, it can be used to determine convergence or divergence much easier than using other techniques in these very specific cases.

Root Test

For the series \( \sum{a_n} \), let \( \displaystyle{L = \lim_{n \to \infty}{\sqrt[n]{|a_n|}}} \).

Three cases are possible depending on the value of L.
\( L < 1 \): The series converges absolutely.
\( L = 1 \): The Root Test is inconclusive.
\( L> 1 \): The series diverges.

When To Use The Root Test

The Root Test is used when you have a function of n that also contains a power with an n. The idea is to remove or change the n in the power. The test itself is fairly straight-forward. You just need some practice using it to know under what conditions it is best to use it. One good way to learn when to use this test is by building example pages as described in the Study Techniques section.

Some Helpful Rules

A powerful rule that will be useful when using the root test is \(\displaystyle{ \lim_{n \to \infty}{\ln(f(n))} = \ln\left( \lim_{n \to \infty}{f(n)} \right) }\)
This is true because the natural log function is continuous. Here are a few additional rules that may be useful.

\(\displaystyle{ \lim_{n \to \infty}{ \sqrt[n]{p}} = 1 }\)

\(p\) is a positive constant

\(\displaystyle{ \lim_{n \to \infty}{\sqrt[n]{n^p}} = 1 }\)

\(p\) is a positive constant

\(\displaystyle{ \lim_{n \to \infty}{\sqrt[n]{n}} = 1 }\)

special case of previous line with \(p=1\)

\(\displaystyle{ \lim_{n \to \infty}{\sqrt[n]{\ln~n}} = 1 }\)

\(\displaystyle{ \lim_{n \to \infty}{\sqrt[n]{n!}} = \infty }\)

We prove the first four rules above as practice problems on the L'Hôpital's Rule page. The last one can be proven using Stirlings Formula.

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, determine whether these series converge or diverge using the root test.

Basic Problems

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{n^2+1}{2n^2+1}\right]^n } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{n^2+1}{2n^2+1}\right]^n } }\)

Final Answer

The series converges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{n^2+1}{2n^2+1}\right]^n } }\)

Solution

264 solution video

video by PatrickJMT

Final Answer

The series converges by the root test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[3^n e^{-n}\right] } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[3^n e^{-n}\right] } }\)

Final Answer

The series diverges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[3^n e^{-n}\right] } }\)

Solution

267 solution video

video by PatrickJMT

Final Answer

The series diverges by the root test.

close solution

\(\displaystyle{ \sum_{k=1}^{\infty}{ \left[\frac{3^k}{(k+1)^k}\right] } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[\frac{3^k}{(k+1)^k}\right] } }\)

Final Answer

The series converges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{k=1}^{\infty}{ \left[\frac{3^k}{(k+1)^k}\right] } }\)

Solution

270 solution video

video by Krista King Math

Final Answer

The series converges by the root test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4n^2+1}{5n^2-8}\right]^n } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4n^2+1}{5n^2-8}\right]^n } }\)

Final Answer

The series converges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[\frac{4n^2+1}{5n^2-8}\right]^n } }\)

Solution

271 solution video

video by MIP4U

Final Answer

The series converges by the root test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{e^{3n}}{n^n} \right] } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{e^{3n}}{n^n} \right] } }\)

Final Answer

The series converges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{e^{3n}}{n^n} \right] } }\)

Solution

273 solution video

video by MIP4U

Final Answer

The series converges by the root test.

close solution

Intermediate Problems

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }\)

Final Answer

The series diverges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^n}{3^{1+3n}} \right] } }\)

Solution

265 solution video

video by PatrickJMT

Final Answer

The series diverges by the root test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n}{n+1} \right]^{n^2} } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n}{n+1} \right]^{n^2} } }\)

Final Answer

The series converges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n}{n+1} \right]^{n^2} } }\)

Solution

266 solution video

video by PatrickJMT

Final Answer

The series converges by the root test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n(n^n)}{3^{n^3+1}} \right] } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n(n^n)}{3^{n^3+1}} \right] } }\)

Final Answer

The series converges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n(n^n)}{3^{n^3+1}} \right] } }\)

Solution

268 solution video

video by PatrickJMT

Final Answer

The series converges by the root test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{[\tan^{-1}(n)]^n} \right] } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{[\tan^{-1}(n)]^n} \right] } }\)

Final Answer

The series converges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{(-1)^n}{[\tan^{-1}(n)]^n} \right] } }\)

Solution

269 solution video

video by PatrickJMT

Final Answer

The series converges by the root test.

close solution

\(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{n-1}}{2^{3+n}} \right] } }\)

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{n-1}}{2^{3+n}} \right] } }\)

Final Answer

The series diverges by the root test.

Problem Statement

Determine convergence or divergence of the series \(\displaystyle{ \sum_{n=1}^{\infty}{ \left[ \frac{n^{n-1}}{2^{3+n}} \right] } }\)

Solution

272 solution video

video by MIP4U

Final Answer

The series diverges by the root test.

close solution
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