Let $r>4$ and $n>1$ be positive integers. Can we simplify this sum:
$$S=\sum_{m=1}^{n}\frac{2m}{r^{m^2}}$$
I have no idea to start.
Answer
Such a sum is related with a Jacobi theta function, hence it cannot be "simplified" too much, but it can be approximated in a quite effective way.
Let $K=\log r$. Then:
$$ S = \frac{1}{K}\sum_{m=1}^{n} 2mK\, e^{-Km^2} $$
can be seen as a Riemann sum, hence:
$$ S \approx \frac{n^2}{K}\int_{0}^{1}2K x\, e^{-K n^2 x^2}\,dx =\frac{1-e^{-K n^2}}{K}.$$
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