I've been working with the series:


From the ratio test it is clear that the series converges for $|x| < 1$, but I'm unable to obtain the sum of the series.

I'm looking for any hint of how to obtain the sum.

Thanks in advance!

  • 4
    $\begingroup$ Differentiate by x, use the geometric sum, integrate, and use that f(0)=0. Result is arctan(x) $\endgroup$ – Ákos Somogyi Aug 25 '15 at 17:11

Let $f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{2n-1}}{2n-1}$. Then we have

$$\begin{align} f'(x)=\sum_{n=1}^{\infty} (-1)^{n+1}x^{2n-2}&=\frac{-1}{x^2}\sum_{n=1}^{\infty} (-x^2)^{n}\\\\ &=\frac{1}{1+x^2} \tag 1 \end{align}$$

Integrating $(1)$ and using $f(0)=0$ reveals that

$$\bbox[5px,border:2px solid #C0A000]{f(x)=\arctan(x)}$$

  • $\begingroup$ (+1) Similar to the approach I was going to take. I was going to note that it was the same as $$\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}$$ and then show that the derivative of that was $\frac1{1+x^2}$ $\endgroup$ – robjohn Aug 25 '15 at 17:46
  • $\begingroup$ @robjohn Thanks! I skipped any discussion of interval of convergence and also omitted discussing the legitimacy of differentiating term by term since it appeared that the OP was already aware of at least the first of these issued. $\endgroup$ – Mark Viola Aug 25 '15 at 17:50

Let $y$ be our sum so:

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

Let's differentiate it to get:

$$ y'=\sum_{n=1}^{\infty}{{(-1)}^{n+1}{x}^{2n-2}}\\ y'=\sum_{n=1}^{\infty}{{(-1)}^{n-1}{x}^{2n-2}}\\ y'=\sum_{n=1}^{\infty}{{(-1)}^{n+1}{(-x^2)}^{n-1}}\\ y'=\frac{1}{1+x^2} $$

Now integrate to get the original sum to get:

$$ y=\arctan{x} +C $$

It is easy to see that for $x=0$ we have $y=0$, so $C=0$, hence the sum equals:

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

  • $\begingroup$ Isn't that just a repeat of what's already posted? $\endgroup$ – Macavity Aug 25 '15 at 17:40
  • $\begingroup$ I was typing my answer at the same time that's why I didn't notice the other answer $\endgroup$ – Oussama Boussif Aug 25 '15 at 17:43
  • 1
    $\begingroup$ (+1) good answer. I've had it happen many times that a similar answer has been posted while I am typing up mine. In fact, it happened here, but I happened to see Dr. MV's answer before I posted. $\endgroup$ – robjohn Aug 25 '15 at 17:49


If $S=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{x^{2n-1}}{2n-1},$

$$i S=\sum_{n=1}^\infty(-1)^{n+1}\dfrac{(ix)^{2n-1}}{2n-1}$$

Now for $-1<y\le1,\ln(1+y)=y-\dfrac{y^2}2+\dfrac{y^3}3-\dfrac{y^4}4+\cdots$



$$\implies2i S=\ln(1+ix)-\ln(1-ix)=\ln\dfrac{1+ix}{1-ix}$$

Let $1=r\cos A,x=r\sin A, x=\tan A$

Now, $$\ln\dfrac{1+ix}{1-ix}=\ln(e^{2iA})=2iA=2i\arctan x$$

  • $\begingroup$ I know what you're trying to show, but the appearance of $x$ and $y$ and the fact that I think the sum for $iS$ looks like it has too many $i^{2n-1}$ in it, makes things a bit hard to follow. $\endgroup$ – robjohn Aug 25 '15 at 18:02

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