$$\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}$$
Euler sum of odd index
- Tolaso J Kos
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Euler sum of odd index
Let $\mathcal{H}_n$ denote the $n$-th harmonic number. Evaluate the sum:
$$\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}$$
$$\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}$$
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Re: Euler sum of odd index
Recall the generating function for harmonic numbers:
$$\sum_{n=1}^\infty\,H_nx^n = - \frac{1}{1-x}\,\ln(1-x).$$
Integrating this yields
$$\sum_{n=1}^\infty\,\frac{H_n}{n+1}x^{n+1} = \frac{1}{2}\,\ln^2(1-x).$$
This can be rewritten as
$$\sum_{n=1}^\infty\,\frac{H_n}{n}x^{n} = \frac{1}{2}\,\ln^2(1-x) + Li_2(x),\qquad(1)$$
where $Li_2$ is the polylogarithm function. Thus, it follows from (1) that
$$\sum_{n=0}^\infty\,\frac{H_{2n+1}}{2n+1}x^{2n+1} = \frac{1}{4}\,\left(\ln^2(1-x)-\ln^2(1+x)\right) + \frac{1}{2}\,(Li_2(x) - Li_2(-x)).\qquad(2)$$
Setting $x = \sqrt{2}/2$ in (2), after some simplification, yields
$$\sum_{n=0}^\infty\,\frac{H_{2n+1}}{2^n(2n+1)} = -\frac{\sqrt{2}}{4}\ln2\ln(3-2\sqrt{2}) + \sum_{n=0}^\infty\,\frac{1}{2^n(2n+1)^2}.$$
Unfortunately, I can not find the exact value of the last series.
$$\sum_{n=1}^\infty\,H_nx^n = - \frac{1}{1-x}\,\ln(1-x).$$
Integrating this yields
$$\sum_{n=1}^\infty\,\frac{H_n}{n+1}x^{n+1} = \frac{1}{2}\,\ln^2(1-x).$$
This can be rewritten as
$$\sum_{n=1}^\infty\,\frac{H_n}{n}x^{n} = \frac{1}{2}\,\ln^2(1-x) + Li_2(x),\qquad(1)$$
where $Li_2$ is the polylogarithm function. Thus, it follows from (1) that
$$\sum_{n=0}^\infty\,\frac{H_{2n+1}}{2n+1}x^{2n+1} = \frac{1}{4}\,\left(\ln^2(1-x)-\ln^2(1+x)\right) + \frac{1}{2}\,(Li_2(x) - Li_2(-x)).\qquad(2)$$
Setting $x = \sqrt{2}/2$ in (2), after some simplification, yields
$$\sum_{n=0}^\infty\,\frac{H_{2n+1}}{2^n(2n+1)} = -\frac{\sqrt{2}}{4}\ln2\ln(3-2\sqrt{2}) + \sum_{n=0}^\infty\,\frac{1}{2^n(2n+1)^2}.$$
Unfortunately, I can not find the exact value of the last series.
- Tolaso J Kos
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Re: Euler sum of odd index
Hello mathofusva. Well I left it the way it was in equation $(2)$ of yours. That is why I said the final expression may contain polylogs.
$$\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)} = - \frac{\sqrt{2}}{4} \ln 2 \ln \left ( 3-2\sqrt{2} \right ) + \frac{1}{4}\Phi \left ( \frac{1}{2}, 2, \frac{1}{2} \right )$$
I am pleased with either formula. More information on Lerch Transcendent may be found here.
That is because Lerch Transcendent does not behave well in this case. See after your simplifications we get that:mathofusva wrote: $$\sum_{n=0}^\infty\,\frac{H_{2n+1}}{2^n(2n+1)} = -\frac{\sqrt{2}}{4}\ln2\ln(3-2\sqrt{2}) + \sum_{n=0}^\infty\,\frac{1}{2^n(2n+1)^2}.$$
Unfortunately, I can not find the exact value of the last series.
$$\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)} = - \frac{\sqrt{2}}{4} \ln 2 \ln \left ( 3-2\sqrt{2} \right ) + \frac{1}{4}\Phi \left ( \frac{1}{2}, 2, \frac{1}{2} \right )$$
I am pleased with either formula. More information on Lerch Transcendent may be found here.
Imagination is much more important than knowledge.
- Tolaso J Kos
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Re: Euler sum of odd index
You know where this came from? Well it arose while playing around with this little fellow:
$$\mathcal{J}=\int_0^1 \frac{\ln (1-x)}{2-x^2} \, {\rm d}x$$
Well,
\begin{align*}
\int_{0}^{1}\frac{\ln (1-x)}{2-x^2} \, {\rm d}x &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^n} \int_{0}^{1}x^{2n} \ln (1-x) \, {\rm d}x \\
&\overset{(*)}{=} -\frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n} \frac{\mathcal{H}_{2n+1}}{2n+1}\\
&=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}
\end{align*}
$(*)$ since $\displaystyle \int_{0}^{1}x^n \ln (1-x) \, {\rm d}x = -\frac{\mathcal{H}_{n+1}}{n+1}$.
$$\mathcal{J}=\int_0^1 \frac{\ln (1-x)}{2-x^2} \, {\rm d}x$$
Well,
\begin{align*}
\int_{0}^{1}\frac{\ln (1-x)}{2-x^2} \, {\rm d}x &=\frac{1}{2}\sum_{n=0}^{\infty} \frac{1}{2^n} \int_{0}^{1}x^{2n} \ln (1-x) \, {\rm d}x \\
&\overset{(*)}{=} -\frac{1}{2} \sum_{n=0}^{\infty} \frac{1}{2^n} \frac{\mathcal{H}_{2n+1}}{2n+1}\\
&=-\frac{1}{2}\sum_{n=0}^{\infty} \frac{\mathcal{H}_{2n+1}}{2^n (2n+1)}
\end{align*}
$(*)$ since $\displaystyle \int_{0}^{1}x^n \ln (1-x) \, {\rm d}x = -\frac{\mathcal{H}_{n+1}}{n+1}$.
Imagination is much more important than knowledge.
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Re: Euler sum of odd index
It seems that there is a closed form for the even index. In fact, we have
$$\sum_{n=1}^\infty\,\frac{H_{2n}}{(2n)\,2^n} = \frac{1}{4}\left(\ln^2\left(\frac{2 -\sqrt{2}}{2}\right)+ \ln^2\left(\frac{2 +\sqrt{2}}{2}\right)\right) + \frac{\pi^2}{48} - \frac{1}{8}\,\ln^22.$$
To this end, as we did above, we have
$$\sum_{n=1}^\infty\,\frac{H_{2n}}{2n}\,x^{2n} = \frac{1}{4}\,(\ln^2(1-x)+ \ln^2(1+x)) + \frac{1}{2}\,(\mbox{Li}_2(x) + \mbox{Li}_2(-x)) \tag{1} $$
In view of the facts that $\mbox{Li}_2(x) + \mbox{Li}_2(-x) = \frac{1}{2}\,\mbox{Li}_2(x^2)$ and
$$\mbox{Li}_2(1/2) = \frac{1}{12}\,\pi^2 - \frac{1}{2}\,\ln^22,$$
Letting $x = 1/\sqrt{2}$ in (1) yields the claimed result.
$$\sum_{n=1}^\infty\,\frac{H_{2n}}{(2n)\,2^n} = \frac{1}{4}\left(\ln^2\left(\frac{2 -\sqrt{2}}{2}\right)+ \ln^2\left(\frac{2 +\sqrt{2}}{2}\right)\right) + \frac{\pi^2}{48} - \frac{1}{8}\,\ln^22.$$
To this end, as we did above, we have
$$\sum_{n=1}^\infty\,\frac{H_{2n}}{2n}\,x^{2n} = \frac{1}{4}\,(\ln^2(1-x)+ \ln^2(1+x)) + \frac{1}{2}\,(\mbox{Li}_2(x) + \mbox{Li}_2(-x)) \tag{1} $$
In view of the facts that $\mbox{Li}_2(x) + \mbox{Li}_2(-x) = \frac{1}{2}\,\mbox{Li}_2(x^2)$ and
$$\mbox{Li}_2(1/2) = \frac{1}{12}\,\pi^2 - \frac{1}{2}\,\ln^22,$$
Letting $x = 1/\sqrt{2}$ in (1) yields the claimed result.
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