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Let $a_n>0$ and $b_n>0$ be two strictly declining sequences such that the series $$\sum_{n=1}^\infty \frac{a_n}{b_n}$$ is convergent. For $\sigma>0$ define $$f^N(\sigma) = \sum_{n=1}^N \frac{a_n}{b_n + \sigma/N}$$ Is it generally true that $\lim_{N \to \infty} f^N(\sigma)$ is independent of $\sigma$ or are there counterexamples?


  1. The answer is trivially true if $\sum \frac{a_n}{b_n^2}$ is convergent as well. In this case $$\left|\frac{d}{d\sigma} f^N(\sigma)\right| = \frac{1}{N}\sum_{n=1}^{N} \frac{a_n}{(b_n+\sigma/N)^2} \leq \frac{1}{N}\sum_{n=1}^N \frac{a_n}{b_n^2} \to 0$$
  2. More interesting is the case of divergent $\sum \frac{a_n}{b_n^2}$, e.g. $a_n = c^{-2n}$ and $b_n = c^{-n}$, or $a_n = 1/n^4$ and $b_n = 1/n^2$. In both these cases $$ \frac{d}{d\sigma} \left.f^N(\sigma)\right|_{\sigma=0} \to 1, $$ but from playing around with Maple and Mathematica I have the suspicion that $\frac{d}{d{\sigma}}f^N(\sigma)$ converges to $0$ for every $\sigma>0$, i.e. $f^N(\sigma)$ becomes non-differentiable in the limit. If that is true it would still allow for the possibility of $f^N(\sigma)$ converging pointwise to a constant.
  3. Eventually I am interested in the case $a_n = n^2I_n(K)^2$ and $b_n=I_n(K)$, where $I_n(K)$ is the modified Bessel function of the first kind.

Any help or pointer to relevant literature is very much appreciated!

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up vote 1 down vote accepted

A very satisfactory answer to this question has been given at MathOverflow:

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