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We know that $$ x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \frac{x^5}{5} - \cdots (-1)^n \frac{x^{n+1}}{n+1} + \cdots = \log (1+x) $$ whenever $-1<x<1$. What can we say if $x=1$?

That is, does the series $$ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots (-1)^n \frac{1}{n+1} + \cdots $$ converge to $\log 2$? If so, how to prove this fact rigorously?

And what about the behavior of the series $$ x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \frac{x^9}{9} - \cdots (-1)^n \frac{x^{2n+1}}{2n+1} + \cdots, $$
which converges to $\arctan x$ for $-1<x<1$, at the points $x = \pm 1$?

Saaqib Mahmood
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2 Answers2

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Daniel's comment already has the solution: since

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

converges (Leibniz series) and the power series

$$\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}n=f(x)$$

converges for $\,|x|<1\,$ to a function $\,f(x)\;$ , then Abel's Theorem gives us

$$\lim_{x\to 1^-}f(x)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}n$$

from here, the sum of the above series indeed is

$$\lim_{x\to 1^-}\log(1+x)=\log2\;$$ since, of course, $\,f(x)=\log(1+x)\;$

DonAntonio
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  • Abel's Theorem tells us that the alternating series at $x=1$ converges, but it doesn't tell us that $f$ is continuous at $x=1$. – robjohn Aug 03 '13 at 09:49
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    @robjohn That's exactly what Abel's theorem tells us. (Radially continuous at least.) – mrf Aug 03 '13 at 09:53
  • Oh, yes it does, @robjohn: that is what's great (and what it is in fact) with this theorem. Well, in fact it says the power series is continuous at $,1,$ from the left. The theorem is easily generalizable for any $,t\in\Bbb R;$ , not necessariyl $,r=1,$ . For ease I'm changing the link I gave. – DonAntonio Aug 03 '13 at 09:53
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    I apologize. I was thinking of Dirichlet's Convergence Test, which is proved similarly (using Abel summation, or summation by parts). Indeed, Abel's Theorem is sufficient. – robjohn Aug 03 '13 at 10:01
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Note that for $x\lt1$ $$ \frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\gt0 $$ and $$ \frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\right) =x^{2k-2}-x^{2k-1}\gt0 $$ Thus, as $x\to1^-$, $$ f(x)=\sum_{k=1}^\infty\left(\frac{x^{2k-1}}{2k-1}-\frac{x^{2k}}{2k}\right) $$ is a positive series, increasing termwise to $$ f(1)=\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right) $$ which converges. Thus, by dominated convergence, $$ \lim_{x\to1^-}f(x)=f(1) $$ The same argument works for $$ g(x)=\sum_{k=1}^\infty\left(\frac{x^{4k-3}}{4k-3}-\frac{x^{4k-1}}{4k-1}\right) $$

robjohn
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