\hspace{-16}\bf{\sum_{n=1}^{\infty}\;\sum_{m=1}^{\infty}\;\frac{1}{m.n.(m+n+1)}}$
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1057this can be done by the method of partial fractions....
when n is 1....
the series is 11.m(m+2).....m going from 1 to ∞....
by the method of partial fractions...this is 12(1m-1m+2)
the sum of the above series is 12...
similarly when is n is 2.....
the series is 16(1m-1m+3) whose sum is 16....
in other words the the resulting double summation is the sum of the series
11.2+12.3+13.4....... upto ∞
and again by the method of partial fractions the sum of this series is 1....
