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I know that $\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^nx^{2n+1}}{2n+1}$ is the power series expansion for $f(x) = \tan^{-1}x$. The interval of convergence for this series is $(-1,1]$.

If I integrate the power series, I have $\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}$. Interval of convergence is $[-1,1]$. The corresponding function can be integrated using integration by parts $x\tan^{-1}x - \dfrac{1}{2}\ln(1+x^2) + C$. Here, I can solve for $C = \dfrac{1}{2}.$

Let's substitute $x=1$, which is part of the interval of convergence, $\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{(2n+1)(2n+2)} =\tan^{-1}1 -\dfrac{1}{2}\ln(2) + \dfrac{1}{2} = \dfrac{\pi}{4} -\dfrac{1}{2}\ln(2) + \dfrac{1}{2}$.

Whats wrong? Im checking using MS Excel and it doesn't seem to get this sum.

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  • $\begingroup$ My edit was to add a missing dollar sign & the period (end-of-sentence) after it.... I was not aware of \dfrac (which, I now see, gives a bigger display than \frac ) so I had to add the dollar to see what it meant. $\endgroup$
    – DanielWainfleet
    Commented Oct 20, 2021 at 15:02
  • $\begingroup$ How did you get $C = \frac12$? This is incorrect. If you let $x = 0$ in the power series, you obtain $0$, so the only possible value of $C$ is $C = 0$. You should obtain $\frac{\pi}{4} - \frac{\ln(2)}{2}$ as the sum. $\endgroup$
    – Angel
    Commented Oct 20, 2021 at 15:58
  • $\begingroup$ Oops, yes, I saw it now. $C=0$, indeed. But my question is this: Now I know that $\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^n x^{2n+2}}{(2n+1)(2n+2)} = x\tan^{-1}x - \dfrac{1}{2}\ln(1+x^2)$. I want to substitute $x=1$, so $\displaystyle\sum_{n=0}^{+\infty} \dfrac{(-1)^n}{(2n+1)(2n+2)} = \tan^{-1}1 - \dfrac{1}{2} \ln 2 $? But I can't seem to arrive at this sum despite having 7000 terms already in Excel. $x=1$ is part of the interval of convergence, so I was expecting to get the sum $\dfrac{\pi}{4} - \dfrac{1}{2}\ln2$, but wasn't able to. $\endgroup$
    – cgo
    Commented Oct 21, 2021 at 1:12

1 Answer 1

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We have

$$(*) \quad\sum_{n=0}^{+\infty} \dfrac{(-1)^nx^{2n+2}}{(2n+1)(2n+2)}=x\tan^{-1}x - \dfrac{1}{2}\ln(1+x^2) + C$$

if and only if $C=0.$ This can be seen with $x=0$ in $(*)$

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