User ingkanit - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:24:51Z http://mathoverflow.net/feeds/user/30573 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118633/series-of-quotients-with-perturbed-denominator Series of quotients with perturbed denominator ingkanit 2013-01-11T14:54:56Z 2013-01-11T18:22:31Z <p>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?</p> <p>Remarks:</p> <ol> <li>The answer is trivially true if $\sum \frac{a_n}{b_n^2}$ is convergent as well. In this case $$\left|\frac{d}{d\sigma} f^N(\sigma)\right| = \frac{1}{N}\sum_{n=1}^{N} \frac{a_n}{(b_n+\sigma/N)^2} \leq \frac{1}{N}\sum_{n=1}^N \frac{a_n}{b_n^2} \to 0$$</li> <li>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.</li> <li>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.</li> <li>It might be related to the Stolz-Cesaro theorem, but I can't figure out how.</li> </ol> <p>Any help or pointer to relevant literature is very much appreciated!</p> http://mathoverflow.net/questions/118633/series-of-quotients-with-perturbed-denominator/118644#118644 Comment by ingkanit ingkanit 2013-01-11T18:21:03Z 2013-01-11T18:21:03Z Thanks! If I understand you correctly the dominated &quot;function&quot; 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 $n &gt; N$. 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!