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Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

$$\sum_{n=1}^{\infty} (-1)^{n-1}\left(\frac{n}{n^2+1}\right)$$

Here's my work:

$b_n = (\dfrac{n}{n^2+1})$

$b_{n+1} = (\dfrac{n+1}{(n+1)^2+1})$

$\lim\limits_{n \to \infty}(\dfrac{n}{n^2+1}) = \lim\limits_{n \to \infty}(\dfrac{1}{n+1/n})=0$

Then I simplified $b_n - b_{n+1}$ in hopes of showing that the sum would be greater than or equal to $0$, but I failed (and erased my work so that's why I haven't included it).

I know the limit of $|b_n|$ is also 0, and I can use that for testing conditional convergence there, but I would run into the same problem for the second half of the test.

I'm having trouble wrapping my head around tests involving absolute values, or more specifically when I have to simplify them.

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This definitely converges by the alternating series test. The AST asks that the unsigned terms decrease and have a limit of 0. In your case, the terms $\frac{n}{n^2+1}$ do exactly that, so it converges.

Now, which flavor of convergence?

If you take absolute values, the resulting series $\sum_n \frac{n}{n^2+1}$ diverges. You can probably get this quickest by limit comparison: terms are on the order of $1/n$. Also, the integral test here is pretty fast because you can see the logarithm.

To apply limit comparison, let's compare $\sum_n \frac{n}{n^2+1}$ to $\sum_n \frac{1}{n}$. Dividing a term in the first by a term in the second gives $$ (\frac{n}{n^2+1})/(\frac{1}{n}) = \frac{n^2}{n^2+1}. $$ Taking the limit gives $L=1$. Since $L>0$, both series "do the same thing." Since $\sum_n \frac{1}{n}$ diverges, so does $\sum_n \frac{n}{n^2+1}$.

Hence it converges conditionally because it converges, but the series of absolute values does not.

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  • $\begingroup$ Could you please show work? Also, your first statement doesn't really help me without the work that led you to that answer. $\endgroup$ – Shea Sep 3 '17 at 0:19
  • $\begingroup$ Maybe someone else can chime in, but I don't know of an easier way of getting convergence than the alternating series test. $\endgroup$ – Randall Sep 3 '17 at 0:20
  • $\begingroup$ I know that is the easiest way to prove it, but I tried that and ran into problems proving it is monotone decreasing. $\endgroup$ – Shea Sep 3 '17 at 0:20
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    $\begingroup$ Oh, that's not too bad. Take the derivative: it's eventually negative. $\endgroup$ – Randall Sep 3 '17 at 0:21
  • $\begingroup$ I never thought about taking the derivative! That's brilliant! I'll try that now. $\endgroup$ – Shea Sep 3 '17 at 0:22
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One can apply the alternating series test directly, but an alternative approach is to take the difference with the series $\sum_{n=1}^\infty (-1)^{n-1}/n$ which is well-known to converge conditionally.

Details: let $a_n=(-1)^{n-1} n/(n^2+1)$, $b_n=(-1)^{n-1}/n$ and $c_n=a_n-b_n$. Then $$c_n=(-1)^{n-1}\left(\frac n{n^2+1}-\frac1n\right) =\frac{(-1)^n}{n(n^2+1)}.$$ Then $\sum_n b_n$ converges conditionally, and $\sum_n c_n$ converges absolutely, so $\sum_na_n=\sum_n(b_n+c_n)$ converges conditionally.

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In at lot of these sort of problems, it comes down to picking good bounds for the terms. For instance, increasing a denominator decreases the size of a fraction (assuming the denominator and numerator are positive, of course). You can increase the denominator by replacing 1 by n. This results in a smaller fraction. So n/(n2+1) < n/(n2+n) . But n/(n2+n)= 1/(n+1), and by the integral test, that diverges. Since the absolute values are larger than a series that diverges the absolute values diverge.

We can also increase the size of the term by decreasing the denominator by 1, getting n/n2 = 1/n.

The next trick it to take pairs of terms. We know that if n is even, then bn = n/(n2+1) < 1/n. And if n is odd, then bn = -n/(n2+1) < -1/(n+1). So if n is even, then bn + bn+1 < 1/n - 1/((n+1)+1)

1/n - 1/((n+1)+1) = 1/n - 1/(n+2) = ((n+2)-n)/n(n+2) = 2/n(n+2)

We can now use the integral test to show this converges. Since the pairs of terms are smaller than a series that converges, it converges (note that although not all the terms in the original series are positive, all of the pairs of terms sum to a positive amount, therefore this test is valid).

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