## 17Calculus Infinite Series - Practice Problem 2440

Infinite Series Practice Problem 2440

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

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.

What We Are Left With

3. We are left with the ratio test, the comparison tests, the integral test and the root test.

Applying The Tests

### Limit Comparison Test

Compare to $$\sum{1/n^2}$$, which is a convergent p-series with $$p = 2$$
$$\displaystyle{a_n=\frac{1}{n^2+4}}$$, $$\displaystyle{t_n=\frac{1}{n^2}}$$

$$\begin{array}{rcl} \displaystyle{ \lim_{n\to\infty}{ \frac{a_n}{t_n} } } & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{1}{n^2+4} \cdot \frac{n^2}{1} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \left[ \frac{n^2}{n^2+4} \frac{1/n^2}{1/n^2} \right] } } \\ & = & \displaystyle{ \lim_{n\to\infty}{ \frac{1}{1+4/n^2} } = 1 } \end{array}$$

Since $$\displaystyle{ 0 \leq \lim_{n\to\infty}{ \frac{a_n}{t_n} } = 1 \lt \infty }$$ and $$\sum{t_n}$$ converges, the series $$\displaystyle{\sum{\frac{1}{n^2+4}}}$$ also converges by the limit comparison test.

### Direct Comparison Test

Compare to $$\sum{1/n^2}$$, which is a convergent p-series with $$p = 2$$
Since the test series $$\sum{1/n^2}$$ is convergent, we assume convergence. This tells us we need to show that the inequality $$0\lt a_n \leq t_n$$ holds.

$$\displaystyle{a_n=\frac{1}{n^2+4}}$$, $$\displaystyle{t_n=\frac{1}{n^2}}$$

It is obvious that $$a_n \gt 0$$, so that part of the inequality holds. Now we need to show that $$a_n \leq t_n$$.

$$\begin{array}{rcl} a_n & \leq & t_n \\ \displaystyle{\frac{1}{n^2+4}} & \leq & \displaystyle{\frac{1}{n^2}} \\ n^2 & \leq & n^2+4 \\ 0 & \leq & 4 \end{array}$$

The last inequality holds for all $$n$$. Therefore since $$0\lt a_n \leq t_n$$ holds and $$\sum{t_n}$$ converges, the series $$\displaystyle{\sum{\frac{1}{n^2+4}}}$$ also converges by the direct comparison test.

### Root Test

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 yes no inconclusive

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

Notes

1. Which test is the 'best' to use for this solution is based on whichever test you are most comfortable using. The root test is the longest and takes the most work to show that the denominator goes to $$1$$. We suggest the limit comparison test or the direct comparison test.
2. Since $$\sum{\left|a_n\right|}$$ converges, the original series converges absolutely by the absolute convergence theorem. This is true by default for all series with positive terms.

Really UNDERSTAND Calculus

 all infinite series topics L'Hopitals Rule

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