Evaluate the double series:
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{ m^2n + mn^2 + 2mn}$$
Solution
1st solution
\begin{align*}
\displaystyle\mathop{\sum}\limits_{n=1}^{\infty}\mathop{\sum}\limits_{m=1}^{\infty}{\frac{1}{m^2n+mn^2+2mn}}&=\mathop{\sum}\limits_{n=1}^{\infty}\mathop{\sum}\limits_{m=1}^{\infty}\Bigl({\frac{1}{n\,(n+2)\,m}-\frac{1}{n\,(n+2)\,(m+n+2)}}\Bigr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}\biggl({\frac{1}{n\,(n+2)}\mathop{\sum}\limits_{m=1}^{\infty}\Bigl({\frac{1}{m}-\frac{1}{m+n+2}}\Bigr)}\biggr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}\biggl({\frac{1}{n\,(n+2)}\mathop{\sum}\limits_{m=0}^{\infty}\Bigl({\frac{1}{m+1}-\frac{1}{m+n+3}}\Bigr)}\biggr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}\Bigl({\frac{1}{n\,(n+2)}\bigl({\psi(n+3)+\gamma}\bigr)}\Bigr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}{\frac{\mathcal{H}_{n+2}}{n\,(n+2)}} \\
&=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n(n+2)}+ \sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} + \sum_{n=1}^{\infty} \frac{1}{n(n+2)^2} \\
&=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n(n+2)}+ \frac{5}{4}- \frac{\pi^2}{12}
\end{align*}
Now we are using the following lemma:
\begin{align*}
\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n(n+2)} &=\int_{0}^{1}x {\rm Li}_2(x) \, {\rm d}x + \frac{1}{2}\int_{0}^{1}x \ln^2 (1-x)\, {\rm d}x \\
&= \left [ \frac{x^2}{2} {\rm Li}_2(x) \right ]_0^1 + \frac{1}{2}\int_{0}^{1} x \ln (1-x) \, {\rm d}x + \frac{1}{2}\int_{0}^{1}x \ln^2 (1-x)\, {\rm d}x\\
&= \frac{\pi^2}{12} -\frac{3}{8} + \frac{7}{8}\\
&= \frac{\pi^2}{12} + \frac{1}{2}
\end{align*}
Returning back we have that:
$$\sum_{n=1}^{\infty}\frac{\mathcal{H}_{n+2}}{n(n+2)}= \frac{7}{4}$$
2st solution [by galactus]
We have successively:
\begin{align*}
\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{m^2n +mn^2+ 2mn} &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{mn \left ( m+n+2 \right )} \\
&= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{mn}\int_{0}^{\infty}e^{-(m+n+2)x} \, {\rm d}x\\
&= \int_{0}^{\infty}e^{-2x} \sum_{n=1}^{\infty} \frac{e^{-nx}}{n} \sum_{m=1}^{\infty}\frac{e^{-mx}}{m} \, {\rm d}x\\
&=\int_{0}^{\infty} e^{-2x} \left ( \sum_{m=1}^{\infty} \frac{e^{-mx}}{m} \right )^2 \, {\rm d}x \\
&= \int_{0}^{\infty} e^{-2x} \ln ^2 (1-e^{-x}) \, {\rm d}x \\
&\overset{u=e^{-x}}{=\! =\! =\!} \int_{0}^{1} u \ln^2 (1-u) \, {\rm d}u \\
&\overset{y=\ln (1-u)}{=\! =\! =\! =\! =\! =\!=\!} \int_{0}^{\infty} y^2 e^{-y} \, {\rm d}y - \int_{0}^{\infty}y^2 e^{-2y} \, {\rm d}y \\
&=\Gamma(3)- \frac{1}{4} \\
&=2 - \frac{1}{4} \\
&=\frac{7}{4}
\end{align*}
$$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{ m^2n + mn^2 + 2mn}$$
Solution
1st solution
\begin{align*}
\displaystyle\mathop{\sum}\limits_{n=1}^{\infty}\mathop{\sum}\limits_{m=1}^{\infty}{\frac{1}{m^2n+mn^2+2mn}}&=\mathop{\sum}\limits_{n=1}^{\infty}\mathop{\sum}\limits_{m=1}^{\infty}\Bigl({\frac{1}{n\,(n+2)\,m}-\frac{1}{n\,(n+2)\,(m+n+2)}}\Bigr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}\biggl({\frac{1}{n\,(n+2)}\mathop{\sum}\limits_{m=1}^{\infty}\Bigl({\frac{1}{m}-\frac{1}{m+n+2}}\Bigr)}\biggr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}\biggl({\frac{1}{n\,(n+2)}\mathop{\sum}\limits_{m=0}^{\infty}\Bigl({\frac{1}{m+1}-\frac{1}{m+n+3}}\Bigr)}\biggr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}\Bigl({\frac{1}{n\,(n+2)}\bigl({\psi(n+3)+\gamma}\bigr)}\Bigr)\\
&=\mathop{\sum}\limits_{n=1}^{\infty}{\frac{\mathcal{H}_{n+2}}{n\,(n+2)}} \\
&=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n(n+2)}+ \sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} + \sum_{n=1}^{\infty} \frac{1}{n(n+2)^2} \\
&=\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n(n+2)}+ \frac{5}{4}- \frac{\pi^2}{12}
\end{align*}
Now we are using the following lemma:
Lemma: It holds that:Thus, from the lemma, we have that:
$$\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n}x^n = {\rm Li}_2(x)+ \frac{1}{2}\ln^2 (1-x)$$
Proof:Simply integrate :
$$\sum_{n=1}^{\infty} \mathcal{H}_n x^{n-1} = -\frac{\ln (1-x)}{x(1-x)}$$
\begin{align*}
\sum_{n=1}^{\infty}\frac{\mathcal{H}_n}{n(n+2)} &=\int_{0}^{1}x {\rm Li}_2(x) \, {\rm d}x + \frac{1}{2}\int_{0}^{1}x \ln^2 (1-x)\, {\rm d}x \\
&= \left [ \frac{x^2}{2} {\rm Li}_2(x) \right ]_0^1 + \frac{1}{2}\int_{0}^{1} x \ln (1-x) \, {\rm d}x + \frac{1}{2}\int_{0}^{1}x \ln^2 (1-x)\, {\rm d}x\\
&= \frac{\pi^2}{12} -\frac{3}{8} + \frac{7}{8}\\
&= \frac{\pi^2}{12} + \frac{1}{2}
\end{align*}
Returning back we have that:
$$\sum_{n=1}^{\infty}\frac{\mathcal{H}_{n+2}}{n(n+2)}= \frac{7}{4}$$
2st solution [by galactus]
We have successively:
\begin{align*}
\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{m^2n +mn^2+ 2mn} &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{mn \left ( m+n+2 \right )} \\
&= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{1}{mn}\int_{0}^{\infty}e^{-(m+n+2)x} \, {\rm d}x\\
&= \int_{0}^{\infty}e^{-2x} \sum_{n=1}^{\infty} \frac{e^{-nx}}{n} \sum_{m=1}^{\infty}\frac{e^{-mx}}{m} \, {\rm d}x\\
&=\int_{0}^{\infty} e^{-2x} \left ( \sum_{m=1}^{\infty} \frac{e^{-mx}}{m} \right )^2 \, {\rm d}x \\
&= \int_{0}^{\infty} e^{-2x} \ln ^2 (1-e^{-x}) \, {\rm d}x \\
&\overset{u=e^{-x}}{=\! =\! =\!} \int_{0}^{1} u \ln^2 (1-u) \, {\rm d}u \\
&\overset{y=\ln (1-u)}{=\! =\! =\! =\! =\! =\!=\!} \int_{0}^{\infty} y^2 e^{-y} \, {\rm d}y - \int_{0}^{\infty}y^2 e^{-2y} \, {\rm d}y \\
&=\Gamma(3)- \frac{1}{4} \\
&=2 - \frac{1}{4} \\
&=\frac{7}{4}
\end{align*}
A generalization may be found here
ReplyDelete\begin{align*}\sum\limits_{n=1}^{\infty} \frac{\mathcal{H}_{n+a}}{n(n+a)} &= -\sum\limits_{n=1}^{\infty} \frac{1}{n}\int_0^{1} x^{n+a-1} \log (1-x)\,dx\\&= \int_0^{1} x^{a-1} \log^{2}(1-x)\,dx\\&= \lim\limits_{b \to 1}\frac{\partial^2}{\partial b^2}\int_0^{1} x^{a-1}(1-x)^{b-1}\,dx \\&= \lim\limits_{b \to 1} \frac{\partial^2}{\partial b^2} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\\&= \lim\limits_{b \to 1}\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\left((\psi_0(b) - \psi_0(a+b))^2 + \psi_1(b) - \psi_1(a+b)\right)\\&= \frac{1}{a}\left((\gamma + \psi_0(a+1))^2 + \frac{\pi^2}{6} - \psi_1(a+1)\right)\\
&=\frac{1}{a} \left ((\mathcal{H}_{a})^2+\mathcal{H}^{(2)}_{a} \right )\end{align*}
since $\mathcal{H}_a = \int_0^1 \frac{1-x^a}{1-x}\,dx = \gamma + \psi_0(1+a)$ and thus by integrating we have that:
$$\int_0^1 x^{a-1} \log (1-x)\,dx = -\frac{\mathcal{H}_a}{a}$$