# Question on Series.

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. :)

• The inner sum is a geometric series, you can calculate explicitly. – André Nicolas Mar 17 '14 at 20:22

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....

• Perfect. This just made me see what I needed to see. Thanks! – user133458 Mar 17 '14 at 20:30

$\sum\limits_{a=2}^{\infty}\sum\limits_{b=2}^{\infty} \frac{1}{a^b}=\sum\limits_{a=2}^{\infty}(a^2-a)^{-1}=\sum\limits_{a=2}^{\infty}((a-1)^{-1}-a^{-1})=(1/1-1/2)+(1/2-1/3)+...=1$.

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.


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