In 'Murty, M. Ram; Murty, V. Kumar, Mean values of derivatives of modular L-series. Ann. of Math. (2) 133 (1991), no. 3, 447–475.', p.456, the author claim the lemma
$\sum_{m=1}^{\infty} \dfrac{1}{m}\exp(-tm^2)=O(\exp(-t))$ $\quad$ if $t \geq 1/2$.
However, I couldn't explain why this holds. The author says that this follows easily by partial summation. How we apply partial summation formula to get this lemma?
Best Answer
I don't think there is a formula here, just write out the first few terms:
$$\sum_{m=1}^\infty \frac{\exp(-tm^2)}{m} = \exp(-t)+ \frac{\exp(-4t)}{2} + \cdots$$
So we can rephrase our question as under what conditions does the first term uniformly dominate the rest of the summation? Rewrite the sum in the following way:
$$\sum_{m=1}^\infty \frac{\exp(-tm^2)}{m} = \exp(-t) + \sum_{m=2}^\infty \frac{\exp(-tm^2)}{m}$$
$$ \leq \exp(-t) + \sum_{m=2}^\infty \frac{\exp(-m)}{m} = \exp(-t) - \log(1-e^{-1}) + 1 = O(\exp(-t))$$
But the inequality step only holds if $tm \geq 1$ for all $m$. The smallest $m$ in the summation is $2$, so it follows that $t\geq \frac{1}{2}$