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 $y$.
Uniform convergence for a summation involving exponentials
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calculus
real-analysis
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2I'd just recognize it as a geometric series... – 2011-10-21
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0But what if say I wanted to use the fact that the series converged uniformly and evaluate the sum $\sum me^{-2\pi my}$? – 2011-10-21
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0what makes you think it converges uniformly? – 2011-10-21