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$$\sum^{\infty}_{n=1} \frac{(-1)^n}{n^2}$$

I know how to prove this one is convergent by alternating test. like Let $b_n = \frac{1}{n^2}$ and limit of this sum is $0$ and also decreasing. However, my professor doesn't allow me to use that I did not learn yet. So I cannot use alternating test(I can use p-series, comparison, ratio test). I don't know how to solve this one..... without alternating test.

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  • $\begingroup$ Consider the largest possible value this could be, i.e., do a bounding analysis. For example, what if you just look at a bound on the magnitude? $\endgroup$ – Michael Mar 31 '17 at 1:19
  • $\begingroup$ Could compare to $\frac{(-1)^n}{n}$ (although you probably used alternating series test for that one too) $\endgroup$ – user12345 Mar 31 '17 at 1:21
  • $\begingroup$ Which tests have you learned so far? $\endgroup$ – mephistolotl Mar 31 '17 at 1:22
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    $\begingroup$ You can use the fact that an absolutely convergent series is conditionally convergent and just use the p-test $\endgroup$ – B.A Mar 31 '17 at 1:22
  • $\begingroup$ Could you show me how to use the fact that "an absolutely convergent series is conditionally convergent" $\endgroup$ – Kwangi Yu Mar 31 '17 at 1:31
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You know about $p$-series, so you know the series $\sum_{n=1}^\infty \frac{1}{n^2}$ converges.

Proving convergence of your alternating series $\sum_{n=1}^\infty \frac{(-1)^n}{n^2}$ is equivalent to proving the sequence of its partial sums, $(S_n)_{n\geq 1}$ is Cauchy.

It will be helpful to compare it to the $p$-series you do know something about; so observe that for any $m\geq 1$ ,$p\geq 0$, $$ \left\lvert\sum_{n=m+1}^{m+p} \frac{(-1)^n}{n^2} \right\rvert \leq \sum_{n=m+1}^{m+p} \left\lvert\frac{(-1)^n}{n^2} \right\rvert = \sum_{n=m+1}^{m+p} \frac{1}{n^2} \tag{1} $$ by the triangle inequality.

So fix $\varepsilon > 0$. Since $\sum_{n=1}^\infty \frac{1}{n^2}$ converges, it is Cauchy, so there exists $m\geq 1$ such that, for all $p\geq 0$, $\sum_{n=m+1}^{m+p} \frac{1}{n^2} \leq \varepsilon$. By the above (1), for the same $m$ and any $p\geq 0$, we have $$ \lvert S_{m+p}-S_m\rvert = \left\lvert\sum_{n=m+1}^{m+p} \frac{(-1)^n}{n^2} \right\rvert \leq \varepsilon. $$

This shows $(S_n)_{n\geq 1}$ is Cauchy, thus convergent.

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  • $\begingroup$ oh... I got it Thank you so much !!!!!!! $\endgroup$ – Kwangi Yu Mar 31 '17 at 1:34
  • $\begingroup$ @KwangiYu You're welcome! Basically, this is a special case of the argument shoing "absolute convergence implies conditional convergence" -- the trick is to use Cauchyness. $\endgroup$ – Clement C. Mar 31 '17 at 1:35
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    $\begingroup$ Assume $\sum_{n}^\infty |a_n|$ converges. Note that we have $0\le a_n+|a_n|\le 2|a_n|$. Then, we can write $$\sum_{n}^N a_n=\sum_{n}^N (a_n+|a_n|)-\sum_{n}^N |a_n|$$which shows that the series of interest is the sum of two convergent series. Done. No need to bring in Cauchy. $\endgroup$ – Mark Viola Mar 31 '17 at 4:30
  • $\begingroup$ @Dr.MV Good point. A "proof from the book" (yet, probably not the most intuitive to a student new to the subject...) $\endgroup$ – Clement C. Mar 31 '17 at 12:35
  • $\begingroup$ CC, I'm not sure which is the less intuitive. I beleve that appealing to Cauchy sequences seem a bit more advanced and less intuitive than appealing to straightforward dominance arguments. $\endgroup$ – Mark Viola Mar 31 '17 at 13:33
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Note that we can write

$$\begin{align} \sum_{n=1}^{2N} \frac{(-1)^n}{n^2}&=\sum_{n=1}^{2N} \frac{1+(-1)^n}{n^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\\\\ &=\sum_{n=1}^N\frac{2}{(2n)^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\\\\ &=\frac12 \sum_{n=1}^N\frac{1}{n^2}-\sum_{n=1}^{2N} \frac{1}{n^2}\tag1 \end{align}$$

Since we know that both of the partial sums on the right-hand side of $(1)$ converge, then the partial sum on the left-hand side must converge also. In fact, we have from $(1)$

$$\sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}=-\frac12 \sum_{n=1}^\infty \frac{1}{n^2}$$

And that completes the proof.


We have a much more general way to approach this problem. We will now show that if a series converges absolutely, then it converges.

To show this, we assume that $\sum_{n=1}^\infty |a_n|$ converges. Now, since $\sum_{n=1}^\infty |a_n|$ converges, then $\sum_{n=1}^\infty 2|a_n|$ converges also.

Note that we have the inequalities $0\le a_n+|a_n|\le 2|a_n|$. Since the sequence $a_n+|a_n|$ is non-negative, then

$$0\le \sum_{n=1}^N(a_n+|a_n|)\le \sum_{n=1}^N2|a_n| \tag 2$$

The sequence of partial sums $S_N=\sum_{n=1}^N(a_n+|a_n|)$ in $(2)$ is monotonically increasing and bounded above by $\sum_{n=1}^\infty 2|a_n|$. Therefore, $S_N$ converges.

Finally, we can write

$$\sum_{n=1}^N a_n=\sum_{n=1}^N (a_n+|a_n|)-\sum_{n=1}^N |a_n| \tag 3$$

Inasmuch as the sequence of partial sums on the right-hand side of $(3)$ converge, the sequence of partial sums on the left-hand side of $(3)$ converge also.

We have just shown that any absolutely convergent sequence also converges.

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  • $\begingroup$ Kwangi, please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Mar 31 '17 at 23:07
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$$\sum \frac{(-1)^n}{n^2} = -1 + \frac{1}{4} - \frac{1}{9} + \frac{1}{16} \cdots = \sum \frac{1}{(2n)^2}- \frac{1}{(2n-1)^2} = \frac{1}{4}\sum\frac{1}{n^2} - \sum \frac{1}{(2n-1)^2}$$

The latter two sums are convergent by $p$ series test.

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