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I had a question that I cannot solve:

So the question is:

Show that $$\sum\limits_{a=2}^{\infty} \sum\limits_{b=2}^{\infty} \frac{1}{a^b}$$

converges to $1$.

So right off the bat, I can see since $b > 1$, this is a convergent p-series. Now I'm stuck, I don't have a strategy to attack this problem. I tried to review series but I still cannot come up with an idea.

If anyone could lead me into the right direction, that would be much appreciated. :)

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  • $\begingroup$ The inner sum is a geometric series, you can calculate explicitly. $\endgroup$ – André Nicolas Mar 17 '14 at 20:22
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Hint: Since $a$ is always greater than $1$, the inner sum is a convergent geometric series,

$$\left({1\over a}\right)^2+\left({1\over a}\right)^3+\left({1\over a}\right)^4+\cdots$$

Find a formula for its sum, and you're off to the races....

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  • $\begingroup$ Perfect. This just made me see what I needed to see. Thanks! $\endgroup$ – user133458 Mar 17 '14 at 20:30
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$\sum\limits_{a=2}^{\infty}\sum\limits_{b=2}^{\infty} \frac{1}{a^b}=\sum\limits_{a=2}^{\infty}(a^2-a)^{-1}=\sum\limits_{a=2}^{\infty}((a-1)^{-1}-a^{-1})=(1/1-1/2)+(1/2-1/3)+...=1$.

In the first step use the formula for a power series knowing that since $a>1,$ $a^{-n}\to0$. In the second step use partial fraction decomposition. The third step is recognizing it as a telescoping series.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\sum_{a = 2}^{\infty}\sum_{b = 2}^{\infty}{1 \over a^{b}}}$ \begin{align} \sum_{a = 2}^{\infty}\sum_{b = 2}^{\infty}{1 \over a^{b}} &=\sum_{a = 2}^{\infty}{1/a^{2} \over 1 - 1/a} =\sum_{a = 2}^{\infty}{1 \over a\pars{a - 1}} =\sum_{a = 2}^{\infty}\pars{{1 \over a - 1} - {1 \over a}} \\[3mm]&=\pars{1 - \half} + \pars{\half - {1 \over 3}} + \pars{{1 \over 3} - {1 \over 4}} + \cdots \end{align}

$$\color{#00f}{\large% \sum_{a = 2}^{\infty}\sum_{b = 2}^{\infty}{1 \over a^{b}} = 1} $$

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