# Convergence of $\sum \limits_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n^2+1}$ [closed]

Test the absolute convergence and convergence of the following series. $$\sum_{n=1}^{\infty}(-1)^{n+1}\frac{n}{n^2+1}$$

Am unable to apply the alternating series test. The book answers it as absolutely convergent. But am not getting the same answer. Kindly help.

Thanks

## closed as off-topic by Empty, choco_addicted, Claude Leibovici, Harish Chandra Rajpoot, Bobson DugnuttMar 10 '16 at 14:50

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• It's not absolutely convergent, and the alternating series test is applicable. Where are you having trouble with it? – user296602 Mar 10 '16 at 4:20
• Note that we have $$\sum_{n=1}^N\frac{n}{n^2+1}\ge\sum_{n=1}^N \frac{1}{2n}=\frac12 \log(N)+O(1)$$So, the series does not converge absolutely. And $\frac{n}{n^2+1}\to 0$ monotonically. So, the alternating series does converge. – Mark Viola Mar 10 '16 at 4:47
• You could also have used the limit comparison test with $\displaystyle\sum_{n=1}^\infty {1\over n}$ to show the lack of absolute convergence. – Christopher Carl Heckman Mar 10 '16 at 5:07

## 3 Answers

The alternating series test is applicable, since $\left|(-1)^{n+1}\frac{n}{n^2+1}\right|$ decreases monotonically.

The series conditionally converges, since $\lim \limits_{n \to \infty} \frac{n}{n^2+1}= \lim \limits_{n \to \infty} \frac{1/n}{1+1/n^2}=0.$

However, the series does not converge absolutely. Using the integral test, we write that $$\displaystyle\int \limits_{0}^{\infty} \frac{xdx}{x^2+1}=\frac{1}{2}\ln (x^2+1)\Big|_1^{\infty}=\infty$$

Since the integral of the absolute values of the terms does not converge, the series does not converge absolutely.

As noted in the comments, you could also use various expansions or comparisons to show that the series doesn't converge.

To apply the alternating series test you need to verify two things: first, that the absolute value of the terms are monotone decreasing, and second, that the terms go to zero in absolute value.

For the first, you need to show that $${n \over n^2 + 1} > {n + 1 \over (n + 1)^2 + 1}$$ This is the same as showing $$n((n + 1)^2 + 1) > (n + 1)(n^2 + 1)$$ Expanding, this is the same as $$n^3 + 2n^2 + 2n > n^3 + n^2 + n + 1$$ Since this is the same as $n^2 + n > 1$, we see that this always holds.

For the second thing you need to show, observe that since $n^2 + 1 > n^2$, one has $$0 < {n \over n^2 + 1} < {n \over n^2}$$ $$= {1 \over n}$$ Since ${\displaystyle {1 \over n}}$ goes to zero as $n$ goes to zero, by the squeeze test so does ${\displaystyle {n \over n^2 + 1}}$. Hence the absolute values of the terms do in fact go to zero.

Consequently, the alternating series test works here and the series converges.

The series does not converge absolutely. For this you can use the limit comparison test with ${\displaystyle {1 \over n}}$ or the integral test.

I had this problem for homework in my AP Calculus BC course.

From what I've been taught, the alternating series test will pretty much always be applicable whenever you see $(-1)^{n}$ or $(-1)^{n+1}$ as part of $a_{n}$.

I approach it in the following manner:

• Let $b_{n}$ equal $a_{n}$ without the alternating component, or in this case $(-1)^{n+1}$.

• Validate two conditions, $\lim_{n\rightarrow \infty } b_{n} = 0$ and $b_{n} > b_{n+1}$ ($b_{n}$ is decreasing).

• If you must determine absolute convergence, use a different test to determine convergence or divergence for $\sum_{n=1}^{\infty} \left | a_{n} \right |$.

So just apply those steps to $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^{2}+1}$.

$b_{n} = \frac{n}{n^{2}+1}$ since $(-1)^{n+1}$ is our alternating component.

$\lim_{n\rightarrow \infty } \frac{n}{n^{2}+1} = 0$ by the exponent rule, where the highest exponent in the denominator is greater than that of the numerator. You can read more about that here.

$b_{n} > b_{n+1}$ can be carried out as follows:

$$\frac{n}{n^{2}+1} > \frac{n+1}{(n+1)^{2}+1}$$

$$\frac{n}{n^{2}+1} > \frac{n+1}{n^{2}+2n+2}$$

$$n^{3}+2n^{2}+2n > n^{3}+n^{2}+n+1$$

So it's pretty clear that $b_{n}$ is decreasing.

$\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^{2}+1}$ is definitely convergent, but to what extent? Of course we can test for absolute convergence by determining the convergence or divergence of $\sum_{n=1}^{\infty} \left | (-1)^{n} \frac{n}{n^{2}+1} \right |$, or $\sum_{n=1}^{\infty} \frac{n}{n^{2}+1}$ since the alternating component $(-1)^{n+1}$ becomes $(1)^{n+1}$ and thus just $1$.

For $\sum_{n=1}^{\infty} \frac{n}{n^{2}+1}$ I'll use the limit comparison test because I like it...

$b_{n}$ is a simplified version of $a_{n}$ and can be determined by simply asking yourself: "What does it look like?"

$b_{n}$ to me looks like $\frac{n}{n^{2}}$, or $\frac{1}{n}$.

According to the limit comparison test, if $\lim_{n\rightarrow \infty } \frac{a_{n}}{b_{n}} > 0$ then $a_{n}$ takes on the properties of $b_{n}$.

Well, $\lim_{n\rightarrow \infty } \frac{a_{n}}{b_{n}}$ is $\lim_{n\rightarrow \infty } \frac{\frac{n}{n^{2}+1}}{\frac{1}{n}}$ or $\lim_{n\rightarrow \infty } \frac{n}{n^{2}+1} * \frac{n}{1}$ or $\lim_{n\rightarrow \infty } \frac{n^{2}}{n^{2}+1} = 1$ by that exponent rule I mentioned earlier. So great, we know now that $a_{n}$ must converge or diverge if $b_{n}$ converges or diverges... but what does it do?

$\sum_{n=1}^{\infty} \frac{1}{n}$ should actually look quite familiar -- it's the friendly and confusing harmonic series. You could validate its divergence by the p-series test, but you should simply know that it diverges.

$b_{n}$ diverges so, by the limit comparison test, $a_{n}$ must also diverge.

Now let's step back and see all that we've just done:

We found that $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^{2}+1}$ does converge by the Alternating Series Test.

We tested $\sum_{n=1}^{\infty} \left | a_{n} \right |$ and found that, by the limit comparison test, it diverged because our $b_{n}$, a harmonic series, diverged.

Therefore, $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n}{n^{2}+1}$ is only conditionally convergent. The summation of $\left | a_{n} \right |$ does not converge, thus it cannot be absolutely convergent.