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I am trying to find the sum of

$$\sum_{k=1}^{\infty}{\frac{(-1)^k}{k}}$$

I've proven that this converges using the Leibniz test, since $a_n > 0$ and $\lim_{n\to\infty}{a_n} = 0$.

I am not sure how to go about summing this series up though. Every example I've seen up to now does some manipulation so that a geometric series pops out. I've been at it for a bit and I don't see how I could convert this to a geometric series to sum it up.

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marked as duplicate by Jack, Math-fun, 2012ssohn, Martin Sleziak, user91500 Dec 1 '16 at 5:42

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    $\begingroup$ It is not enough $\;a_n>0\,,\,a_n\rightarrow 0\;$ . It must be that $\;a_n\;$ is monotone descending, otherwise the series may diverge. $\endgroup$ – DonAntonio Nov 30 '16 at 8:37
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Using the Taylor series for the natural logarithm, $$ \ln(x+1)=\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n} $$ for $-1<x\le 1$. Abel's theorem guarantees that $$ \ln2=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}. $$ Hence, $$\sum_{n=1}^\infty\frac{(-1)^{n}}{n}=-\ln2. $$

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  • $\begingroup$ Could anyone explain the reason for the downvote?.. $\endgroup$ – Cm7F7Bb Nov 30 '16 at 7:57
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    $\begingroup$ The answer is making no connection to the geometric series. And dropping ready-made Taylor is tantamount to saying "this sum is known to be $-\ln 2$". $\endgroup$ – Yves Daoust Nov 30 '16 at 8:05
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    $\begingroup$ @YvesDaoust Do the answers have to make a connection to the geometric series? No, it does not say so in the question. I agree with the remark about the Taylor series, but it is not a reason for a downvote. In any case, we still need to know why we can set $x$ equal to $1$ and I explained that. $\endgroup$ – Cm7F7Bb Nov 30 '16 at 8:24
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    $\begingroup$ Completely agree with Cm: no need to connect with geometric series and the point is this answer uses Abel's theorem to susbtitute $\;x=1\;$ . Taylor series are pretty elementary stuff, though perhaps geometric series are even more elementary. Yet the other answer also uses as "high level" basis calculus when it assumes a power series can be integrated and even without mentioning anything about its convergence radius. $\endgroup$ – DonAntonio Nov 30 '16 at 8:41
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    $\begingroup$ @DonAntonio: Am I crazy ? "Every example I've seen up to now does some manipulation so that a geometric series pops out. I've been at it for a bit and I don't see how I could convert this to a geometric series to sum it up." $\endgroup$ – Yves Daoust Nov 30 '16 at 10:15
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You can find the sum using only basic calculus -- without invoking termwise integration or differentiation of infinite Taylor series or Abel's theorem.

Begin with the finite geometric sum

$$\sum_{k=0}^{n-1} (-x)^k = \frac{1 - (-x)^n}{1+x}.$$

Rearranging, we have

$$\sum_{k=0}^{n-1} (-1)^kx^k - \frac{1}{1+x}= \frac{(-1)^nx^n}{1+x}.$$

Integrate both sides over $[0,1]$, noting that termwise integration of the finite sum requires no special justification,

$$\sum_{k=0}^{n-1} (-1)^k\int_0^1x^k \, dx - \int_0^1\frac{1}{1+x} \, dx= \int_0^1\frac{(-1)^nx^n}{1+x} \, dx.$$

Hence,

$$\sum_{k=0}^{n-1} \frac{(-1)^k}{k+1} - \ln 2= (-1)^n\int_0^1\frac{x^n}{1+x} \, dx,$$

and

$$0 \leqslant \left|\sum_{k=1}^{n} \frac{(-1)^{k+1}}{k} - \ln 2 \right| = \left|\int_0^1\frac{x^n}{1+x} \, dx\right| \\ \leqslant \int_0^1\left|\frac{x^n}{1+x}\right| \, dx \leqslant \int_0^1 x^n \, dx = \frac{1}{n+1}.$$

By the squeeze theorem, it follows that

$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \ln 2.$$

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Since an alternating sum is the difference of the non-alternating sum and twice the sum of the even terms, we get $$ \begin{align} \sum_{k=1}^{2n}\frac{(-1)^{k-1}}k &=\sum_{k=1}^{2n}\frac1k-2\sum_{k=1}^n\frac1{2k}\\ &=\sum_{k=1}^{2n}\frac1k-\sum_{k=1}^n\frac1k\\ &=\sum_{k=n+1}^{2n}\frac1k\tag{1} \end{align} $$ By comparison to an integral, we have $$ \log\left(\frac{2n+1}{n+1}\right)=\int_{n+1}^{2n+1}\frac1x\,\mathrm{d}x\le\sum_{k=n+1}^{2n}\frac1k\le\int_n^{2n}\frac1x\,\mathrm{d}x=\log(2)\tag{2} $$ Apply the Squeeze Theorem to $(2)$ and incorporate $(1)$ and we get $$ \sum_{k=1}^\infty\frac{(-1)^{k-1}}k=\log(2)\tag{3} $$

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Hint:

The geometric series is very very close.

Replace $-1$ by $x$ and derive term-wise:

$$\left(\sum_{k=1}^\infty\frac{x^k}k\right)'=\sum_{k=1}^\infty\left(\frac{x^k}k\right)'=\sum_{k=1}^\infty x^{k-1}.$$

Now you can use the geometric series formula and integrate from $0$ to $-1$ to get the solution

$$-\left.\ln(1-x)\right|_0^{-1}.$$

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  • $\begingroup$ Oh, so you recognised this as a power series so you could derive and integrate it? Could you just briefly explain why we integrate only from $0$ to $-1$? $\endgroup$ – Pavlin Nov 30 '16 at 8:14
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    $\begingroup$ How did you justify differentiation term by term? $\endgroup$ – Vim Nov 30 '16 at 15:21
  • $\begingroup$ @vim: I have turned the answer to a hint. $\endgroup$ – Yves Daoust Nov 30 '16 at 15:27

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