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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?

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!

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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.

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  • $\begingroup$ Thanks! If I understand you correctly the dominated "function" in this case is defined by $g_{N,n} = a_n/(b_n+\sigma/N)$ for $n \leq N$ and $g_{N,n} = 0$ for $n > N$. Then $\sum_{n=1}^\infty g_{N,n} = f^N(\sigma)$ but also $g_{N,n} \to a_n/b_n$ pointwise. Since $g_{N,n}$ is dominated by $a_n/b_n$ the convergence follows by dominated convergence wrt to the counting measure. Very helpful! $\endgroup$ – ingkanit Jan 11 '13 at 18:21
  • $\begingroup$ Exact, it's the dominated convergence wrt to the counting measure (that one can state and prove directly, of course, without measure theory) $\endgroup$ – Pietro Majer Jan 11 '13 at 18:26

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