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Is it true that the series $$ \sum_{n = 1}^{\infty} \frac{\alpha \log n}{1 + (n + \alpha)^2} $$ is uniformly bounded for every $\alpha \geq 0$, i.e., does there exists a constant $C$ such that the series is bounded by $C$ for every $\alpha \geq 0$? I tried to show that it is unbounded, but I always ended up with a constant as a lower bound instead of getting something in terms of $\alpha$. I believe that $\log$ is redundant here, not playing any particular role in the uniform boundedness due to the exponent $2$ in the denominator.

Any hints or suggestions will be greatly appreciated.

Epsilon-Delta
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1 Answers1

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If $\alpha $ is a positive integer then $$ \frac{\alpha \log n}{1 + (n + \alpha)^2} \ge \frac{\alpha \log \alpha}{1+9 \alpha^2} $$ for $\alpha < n \le 2 \alpha$, and the sum has exactly $\alpha$ terms in that range. It follows that

$$ \sum_{n=1}^\infty \frac{\alpha \log n}{1 + (n + \alpha)^2} \ge \frac{\alpha^2 \log \alpha}{1+9 \alpha^2} \ge \frac{1}{10} \log \alpha \, . $$ This proves that the sum is not uniformly bounded for $\alpha \ge 0$.

Martin R
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  • Thank you Martin. I actually wanted to find a bound (up to a multiplicative constant, big O I mean) for the series $\sum_{n = 1}^{\infty} \frac{\log n}{(n + \alpha)^2}$. I thought it might be $1/ \alpha$ but it turns out that it is actually $\log \alpha / \alpha$, which is sufficient for my purposes. – Epsilon-Delta Jan 11 '24 at 13:28