## 17Calculus Infinite Series - Practice Problem 2441

Infinite Series Practice Problem 2441

$$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2-1}{n^3+4}}}$$

Determine the convergence or divergence of the series.
If the series converges,
- determine the value the series converges to, if possible; and
- determine if the series converges absolutely or conditionally.

Choosing Which Tests To Try

Rejecting the Obvious

1. We can tell right away that this is not an alternating series, telescoping series or one of the other special series. So we will not be able to determine what the series converges to.
2. We can tell that the limit of $$a_n$$ is zero. So the divergence test will not help us.
3. The integral is quite complicated, so we will not try the integral test unless all the other tests fail.
4. There are no factors with n in the powers. So we will put the root test aside and come back to it only if the other tests fail. We would probably try this one before the integral test.

What We Are Left With

5. We are left with the ratio test and the comparison tests.

Applying The Tests

### Limit Comparison Test

Use the test series $$\sum{1/n}$$, a divergent p-series.
$$\begin{array}{rcl} \displaystyle{ \lim_{n\to\infty}{ \left| \frac{a_n}{t_n} \right| } } & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^2-1}{n^3+4} \frac{n}{1} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^3-n}{n^3+4} \frac{1/n^3}{1/n^3} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{1-1/n^2}{1+4/n^3} \right] } = 1} \\ \end{array}$$
Since the limit is finite and positive, both series either converge or diverge. Since the test series diverges, so does the original series by the limit comparison test.

### Direct Comparison Test

For this question $$\displaystyle{ a_n = \frac{n^2-1}{n^3+4} }$$. If we think about what happens when n is very large, the $$n^2$$ term dominates the 1 in the numerator and the $$n^3$$ term dominates the 4 in the denominator. When that happens we are very close to $$n^2/n^3 = 1/n$$. So let's apply the direct comparision test using $$t_n = 1/n$$. Since $$\sum{1/n}$$ is a divergent p-series, we will assume the series $$\sum{a_n}$$ diverges also. Using the assumption, the inequality we need to prove is $$0 \lt t_n \leq a_n$$.
It is obvious that $$0 \lt t_n$$, so we will focus on showing that $$t_n \leq a_n$$.
$$\begin{array}{rcl} t_n & \leq & a_n \\ \displaystyle{ \frac{1}{n} } & \leq & \displaystyle{ \frac{n^2-1}{n^3+4} } \\ n^3+4 & \leq & n(n^2-1) \\ n^3+4 & \leq & n^3-n \\ 4 & \leq & -n \\ -4 & \geq & n \end{array}$$
This last inequality does not hold. So we need another test series.
At this point, we will still assume that the series diverges. So we need a divergent series to compare it to. To find one, we plotted the functions $$f(x)=(x^2-1)/(x^3+4)$$ and $$g(x)=1/x$$ on the same set of axes. Sure enough, g(x) is larger than f(x) which confirms our work above. Our test series needs to be less than or equal to our original series. So we need a divergent series that is smaller than $$1/n$$.
To make $$1/n$$ smaller, we can make the denominator larger, so we looked at $$1/(n+1)$$. Plotting $$h(x)=1/(x+1)$$ on the same set of axes as f(x) and g(x), we can zoom in and see that, sure enough, h(x) is smaller than f(x). But is $$\sum{1/(n+1)}$$ still a divergent series? We can use the limit comparison test as follows to find out.
$$\begin{array}{rcl} \displaystyle{ \lim_{n\to\infty}{\frac{t_n}{a_n}} } & = & \displaystyle{ \lim_{n\to\infty}{ \frac{1}{n+1} \frac{n}{1} } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \frac{n}{n+1} } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \frac{1}{1+1/n} } = 1 } \end{array}$$
Since the limit is finite and positive, then both series either converge or diverge. Since the test series diverges, so does $$\sum{1/(n+1)}$$.
So now for the direct comparison test, we have a divergent test series $$t_n=1/(n+1)$$ and we need to show that $$0 \lt t_n \leq a_n$$. As before $$0 \lt t_n$$, so we focus on $$t_n \leq a_n$$.
$$\begin{array}{rcl} t_n & \leq & a_n \\ \displaystyle{ \frac{1}{n+1} } & \leq & \displaystyle{ \frac{n^2-1}{n^3+4} } \\ n^3+4 & \leq & (n+1)(n^2-1) \\ n^3+4 & \leq & n^3-n+n^2-1 \\ 4 & \leq & n^2-n-1 \\ 0 & \geq & n^2-n-5 \end{array}$$
Can we find an N such that for $$n \gt N$$, $$n^2-n-5 \geq 0$$? Yes we can.
Notice that $$y=x^2-x-5$$ is a parabola opening up. So we need to find where the graph crosses the x-axis and choose N larger than that. Completing the square gives us $$x=1/2\pm\sqrt{21/4}$$. We don't care about the negative value. The positive value is about 2.79. So we choose N=3. So the inequality $$0 \geq n^2-n-5$$ holds for $$n \gt 3$$ and therefore the series diverges.

test/series works? notes no limit is zero, so test is inconclusive no not a p-series no not a geometric series no not an alternating series no not a telescoping series no inconclusive yes best test yes no integral too complex

The series $$\displaystyle{ \sum_{n=1}^{\infty}{\frac{n^2-1}{n^3+4}}}$$ diverges.

Really UNDERSTAND Calculus

 all infinite series topics L'Hopitals Rule

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