Let \( \mathcal{H}_n \) denote the \( n \) th harmonic number. Evaluate the sum:
$$ \mathcal{S}=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2} $$
Answer: \( \displaystyle \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2}=2\zeta(3) \)
Proof:
It is well known that the \( n \) th harmonic number has an integral representantion, which is:
$$\mathcal{H}_n=\int_{0}^{1}\frac{1-x^n}{1-x}\,{\rm d}x$$
Hence the sum can be re-written as:
$$ \begin{aligned}
\mathcal{S}=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2} &=\sum_{n=1}^{\infty}\frac{1}{n^2}\int_{0}^{1}\frac{1-x^n}{1-x}\,{\rm d}x \\
&= \int_{0}^{1}\frac{1}{1-x}\sum_{n=1}^{\infty}\left ( \frac{1}{n^2}-\frac{x^n}{n^2} \right )\,{\rm d}x\\
&=\int_{0}^{1}\frac{\frac{\pi^2}{6}-{\rm Li_2}(x)}{1-x}\,{\rm d}x \\
&=\require{cancel} \cancelto{0}{ -\ln(1-x)\left ( \frac{\pi^2}{6}-{\rm Li_2}(x) \right )\bigg|_0^1}+\int_{0}^{1}\frac{\ln^2 (1-x)}{x}\,{\rm d}x\\
&\overset{u=1-x}{=\! =\! =\! =\!}\int_{0}^{1}\frac{\ln^2 u}{1-u}\,{\rm d}u \\
&= \int_{0}^{1}\ln^2 x\sum_{k=0}^{\infty}x^k\,{\rm d}x\\
&= \sum_{k=0}^{\infty}\int_{0}^{1}x^k \ln^2 x \,{\rm d}x\\
&= \cdots \\
&=2\sum_{k=0}^{\infty}\frac{1}{\left ( k+1 \right )^3}=2\zeta(3)
\end{aligned}$$
The result follows and the exercise comes to an end.
$$ \mathcal{S}=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2} $$
Answer: \( \displaystyle \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2}=2\zeta(3) \)
Proof:
It is well known that the \( n \) th harmonic number has an integral representantion, which is:
$$\mathcal{H}_n=\int_{0}^{1}\frac{1-x^n}{1-x}\,{\rm d}x$$
Hence the sum can be re-written as:
$$ \begin{aligned}
\mathcal{S}=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2} &=\sum_{n=1}^{\infty}\frac{1}{n^2}\int_{0}^{1}\frac{1-x^n}{1-x}\,{\rm d}x \\
&= \int_{0}^{1}\frac{1}{1-x}\sum_{n=1}^{\infty}\left ( \frac{1}{n^2}-\frac{x^n}{n^2} \right )\,{\rm d}x\\
&=\int_{0}^{1}\frac{\frac{\pi^2}{6}-{\rm Li_2}(x)}{1-x}\,{\rm d}x \\
&=\require{cancel} \cancelto{0}{ -\ln(1-x)\left ( \frac{\pi^2}{6}-{\rm Li_2}(x) \right )\bigg|_0^1}+\int_{0}^{1}\frac{\ln^2 (1-x)}{x}\,{\rm d}x\\
&\overset{u=1-x}{=\! =\! =\! =\!}\int_{0}^{1}\frac{\ln^2 u}{1-u}\,{\rm d}u \\
&= \int_{0}^{1}\ln^2 x\sum_{k=0}^{\infty}x^k\,{\rm d}x\\
&= \sum_{k=0}^{\infty}\int_{0}^{1}x^k \ln^2 x \,{\rm d}x\\
&= \cdots \\
&=2\sum_{k=0}^{\infty}\frac{1}{\left ( k+1 \right )^3}=2\zeta(3)
\end{aligned}$$
The result follows and the exercise comes to an end.
A comment
ReplyDeleteEuler proved in $1775$ that the following formula holds:
$$2\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^q}= \left ( q+2 \right )\zeta(q+1)- \sum_{m=1}^{q-2}\zeta(m+1)\zeta(q-m) $$
Let us verify this example for the above Euler sum. We have $q=2$ in this case:
$$2\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2}= 4\zeta(3)- 0 \Leftrightarrow \sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n^2}=2\zeta(3)$$
which agrees to what we have actually gotten above.