MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.
  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!

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
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! – ingkanit Jan 11 '13 at 18:21
Exact, it's the dominated convergence wrt to the counting measure (that one can state and prove directly, of course, without measure theory) – Pietro Majer Jan 11 '13 at 18:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.