Find the sum of the iterated series: sum from m=2

Find the sum of the iterated series: $$ \sum_{m=2}^{\infty} \sum_{k=2}^{\infty} \frac1{m^k}. $$

Explanation of Method

For each fixed $m\ge2$, the inner sum over $k$ is a geometric series with ratio $1/m<1$, summing in closed form to $1/[m(m-1)]$. The outer sum over $m$ is then a classical telescoping series via partial fractions, collapsing to a simple limit.

Sum the inner geometric series over $k$ for fixed $m$.
Decompose the resulting $1/[m(m-1)]$ via partial fractions.
Evaluate the outer telescoping sum over $m$.

Step 1 · Inner geometric sum

For fixed $m\ge2$: $$ \sum_{k=2}^{\infty}\frac1{m^k} = \frac{1/m^2}{1-1/m} = \frac1{m(m-1)}. $$

Step 2 · Partial fractions

$\dfrac1{m(m-1)}=\dfrac1{m-1}-\dfrac1m$.

Step 3 · Telescoping outer sum

$$ \sum_{m=2}^{\infty}\left(\frac1{m-1}-\frac1m\right) = \lim_{N\to\infty}\left(1-\frac1N\right) = 1. $$ Verified: partial sums $S_2=\frac12$, $S_3=\frac23$, $S_4=\frac34$, matching $1-1/N$ exactly.

Summary Table

Step	Result
Inner sum (fixed $m$)	$1/[m(m-1)]$
Outer sum	Telescopes to $1$

Additional Notes

This iterated series is a nested version of the elementary telescoping-series technique seen throughout this sequence (e.g. Problem 141), combined with the geometric series formula; both individually elementary tools compose here into a clean closed-form answer.
Since all terms are positive, Tonelli's theorem guarantees the sum is independent of whether $m$ or $k$ is summed first, unlike the delicate signed double series of Problem 150 — a useful point of contrast between this problem and that one.
This construction is a special case of the general double-series identity $\sum_{m\ge2}\sum_{k\ge2}x^k/m^k$-type sums; the specific choice $x=1$ (i.e. summing the reciprocal powers of $m$ directly) is what allows the final telescoping simplification to a clean integer.

Result

The iterated series sums to:

$$ \sum_{m=2}^{\infty}\sum_{k=2}^{\infty}\frac1{m^k} = 1. $$