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?


Remarks:


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.
4. It might be related to the Stolz-Cesaro theorem, but I can't figure out how.


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



Series of quotients with perturbed denominator
Since $\sum\_ {n=1}^\infty \frac{a\_n}{b\_n } < \infty$ and $0 \le \frac{a\_n}{b\_n + \sigma/N}\le \frac{a\_n}{b\_n} $, we have that $\sum\_{n=1}^N \frac{a\_n}{b\_n + \sigma/N} \to \sum\_ {n=1}^\infty \frac{a\_n}{b\_n }$ as $N\to\infty$, just by dominated convergence.


