روش اول:
$$m^2n+mn^2+2mn=mn(m+n+2)$$
$$, \frac{1}{m(m+n+2)}=\frac{1}{n+2}(\frac{1}{m}- \frac{1}{m+n+2} )$$
$$ \Rightarrow \sum_{m=1}^ \infty \frac{1}{m(m+n+2)}=\frac{1}{n+2}\sum_{m=1}^ \infty(\frac{1}{m}- \frac{1}{m+n+2} )$$
$$=\frac{1}{n+2}(\sum_{m=1}^ \infty(\frac{1}{m}- \frac{1}{m+1})+\sum_{m=1}^ \infty(\frac{1}{m+1}- \frac{1}{m+2})+...\sum_{m=1}^ \infty(\frac{1}{m+n+1}- \frac{1}{m+n+2}))$$
$$=\frac{1}{n+2}( \frac{1}{1} + \frac{1}{2} +...+ \frac{1}{n+2} )$$
$$= \frac{H_{n+2}}{n+2} $$
$$ \Rightarrow \sum_{n=1}^ \infty \sum_{m=1}^ \infty \frac{1}{m^2+mn^2+2mn}=\sum_{n=1}^ \infty(\sum_{m=1}^ \infty \frac{1}{mn(m+n+2)} )$$
$$=\sum_{n=1}^ \infty \frac{1}{n} (\sum_{m=1}^ \infty \frac{1}{m(m+n+2)} )$$
$$=\sum_{n=1}^ \infty \frac{H_{n+2}}{n(n+2)}$$
$$=\frac{1}{2}\sum_{n=1}^ \infty H_{n+2}( \frac{1}{n} - \frac{1}{n+2} )$$
$$=\frac{1}{2}\sum_{n=1}^ \infty ( \frac{H_n+ \frac{1}{n+1} + \frac{1}{n+2} }{n} - \frac{H_{n+2}}{n+2} )$$
$$=\frac{1}{2}(\sum_{n=1}^ \infty \frac{H_n}{n}-\sum_{n=1}^ \infty \frac{H_{n+2}}{n+2}+\sum_{n=1}^ \infty \frac{1}{n(n+1)}+\sum_{n=1}^ \infty \frac{1}{n(n+2)})$$
$$=\frac{1}{2}(\sum_{n=1}^ \infty \frac{H_n}{n}-\sum_{n=3}^ \infty \frac{H_{n}}{n}+\sum_{n=1}^ \infty (\frac{1}{n}- \frac{1}{n+1} )+ \frac{1}{2} \sum_{n=1}^ \infty (\frac{1}{n}- \frac{1}{n+2}))$$
$$=\frac{1}{2}( \frac{H_1}{1} + \frac{H_2}{2} + \frac{1}{1} + \frac{1}{2} \sum_{n=1}^ \infty (\frac{1}{n}- \frac{1}{n+1})+ \frac{1}{2} \sum_{n=1}^ \infty (\frac{1}{n+1}- \frac{1}{n+2}))$$
$$=\frac{1}{2}(1+ \frac{1}{2} + \frac{1}{4} +1+ \frac{1}{2} + \frac{1}{4} )$$
$$=\frac{7}{4}$$
روش دوم (صرفن راهنمایی):
$$\frac{1}{m+n+2}= \int_0^1x^{m+n+1}$$
$$,\sum_{k=1}^ \infty \frac{k^k}{k}=-Ln(1-x),0<x<1$$
و از جابجایی سیگما و انتگرال استفاده کنید.
$\Box$
