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For $y > 0$, why does the series $\sum_{m = 1}^{\infty} e^{-2\pi my}$ converge uniformly in $y$? Can I just use the following argument? For $m$ large enough, there exists a constant $C$ independent of $y$ such that $e^{-2\pi my} \leq m^{-2}$. Then as $\sum m^{-2} < \infty$, $\sum_{m = 1}^{\infty} e^{-2\pi my}$ converges uniformly in $y$.

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I'd just recognize it as a geometric series... –  J. M. Oct 21 '11 at 17:10
    
But what if say I wanted to use the fact that the series converged uniformly and evaluate the sum $\sum me^{-2\pi my}$? –  23894 Oct 21 '11 at 17:15
    
what makes you think it converges uniformly? –  Thomas Andrews Oct 21 '11 at 17:22

1 Answer 1

Let $g(y)=e^{-2\pi y}$. Then the partial sum of your series is:

$$f_n(y) = \sum_{m=0}^n g(y)^m = \frac{1-g(y)^{n+1}}{1-g(y)}$$

For $y>0$, $0<g(y)<1$, so this coverges to $f(y)=\frac{1}{1-g(y)}$

Now, to prove uniform convergence, you need for each $\epsilon>0$ an $N$ such that if $n>N$, $\forall y: |f(y)-f_n(y)|<\epsilon$

But $f(y) - f_n(y) = \frac{g(y)^{n+1}}{1-g(y)}$

As $y\rightarrow 0$, $g(y)\rightarrow 1$, so $|f(y)-f_n(y)|$ approaches infinity. In particular, then, for any $n$, there is always a $y$ such that $|f(y)-f_n(y)|>\epsilon$.

Therefore, this doesn't converge uniformly in y.

Note: You can actually show that it cannot be uniformly convergent just by noting that $f(y)$ is unbounded on $(0,+\infty)$, and that each $f_n$ is bounded above (by $n$.)

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