mathimatikoi.orghttps://www.mathimatikoi.org/forum/ A generating function involving harmonic number of even indexhttps://www.mathimatikoi.org/forum/viewtopic.php?f=27&t=1189 Page 1 of 1

 Author: Tolaso J Kos [ Wed Jun 14, 2017 8:33 am ] Post subject: A generating function involving harmonic number of even index Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Prove that forall $|x|<1$ it holds that$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_{2n} x^{2n+1}}{2n+1} = \frac{\arctan x \log (1+x^2)}{2}$

 Author: mathofusva [ Tue Jun 27, 2017 6:26 pm ] Post subject: Re: A generating function involving harmonic number of even index Let the proposed series be $f(x)$. Then$$f'(x) = \sum_{n=1}^\infty (-1)^{n-1}H_{2n}x^{2n} = - \sum_{n=1}^\infty H_{2n}(-x^2)^n.$$Recall that$$\sum_{n=1}^\infty H_nz^n = - \frac{\ln(1-z)}{1-z}.$$It follows that\begin{eqnarray*}\sum_{n=1}^\infty H_{2n}z^{2n} & = & \frac{1}{2}\,(\sum_{n=1}^\infty H_nz^n + \sum_{n=1}^\infty H_n(-z)^n)\\& = & - \frac{1}{2}\left(\frac{\ln(1-z)}{1-z} + \frac{\ln(1+z)}{1+z}\right).\end{eqnarray*}Setting $z = ix$ gives\begin{eqnarray*}\sum_{n=1}^\infty H_{2n}(-x^2)^n & = & - \frac{1}{2}\left(\frac{\ln(1-ix)}{1-ix} + \frac{\ln(1+ix)}{1+ix}\right)\\& = & - \frac{1}{2}\,\frac{2x\arctan x+ \ln(1+x^2)}{1 + x^2},\end{eqnarray*}where we have used $\ln(1\pm ix) = \frac{1}{2}\ln(1 + x^2) \pm i\arctan x$. Thus,$$f'(x) = \frac{1}{2}\,\frac{2x\arctan x+ \ln(1+x^2)}{1 + x^2}.$$Integrating with respect to $x$ yields$$f(x) = \frac{1}{2}\,\arctan x\ln(1+x^2)$$as desired.

 Page 1 of 1 All times are UTC [ DST ] Powered by phpBB® Forum Software © phpBB Grouphttps://www.phpbb.com/