I have a problem to determine convergence (sum over n). $$\sum_{n=0}^\infty \dfrac {a\left( a+1^{p}\right) \ldots \left( a+n^{p}\right) }{b\left( b+1^{p}\right) \ldots \left( b+n^{p}\right) }$$where $a<b, a>0,b>0$.

I have concluded convergence for $p\leq0$ by comparing it to a constructed geometric sequence, as well as for $p=1$, using comparison test with $n^{a-b}$. But I can not use similar methods for $p>1$ and $0<p<1$.

I have some thoughts for the two parts:

When $p>1$, it seems that the limit of each term is not $0$. If the limit could be evaluated, then the divergence can be proved. My method for $p=1$ is to use the Euler Product of the gamma function, but the $p$ power makes it impossible to use this method. I am wondering if there is any kind of generalization of gamma function that is of this form.

when $0<p<1$, I compared it to the case of $p=1$, that could at least tell it converges in the range when $p=1$ converge. But it is inconclusive for the parts remaining.

Any help or hints would be appreciated.

  • $\begingroup$ Is the summation over $n$ ? $\endgroup$ – Claude Leibovici May 1 '15 at 14:08
  • $\begingroup$ @ClaudeLeibovici Yes $\endgroup$ – William Riddle May 1 '15 at 14:10
  • $\begingroup$ Let $a_n(a,p)=a\left( a+1^{p}\right) \ldots \left( a+n^{p}\right) $ we have: $\ln(a_n(a,p))=\sum_{i=1}^{n}\ln(a+i^p)$ and this can be compared to: $$\int_{0}^{n}\ln(1+x^p)dx $$ $\endgroup$ – Elaqqad May 1 '15 at 14:52
  • $\begingroup$ @Elaqqad Thanks, but when I tried this, it turned out to be $$\sum ^{n}_{i=1}\ln \left( a+i^{p}\right)-\sum ^{n}_{i=1}\ln \left( b+i^{p}\right)$$ for a single term, the comparison can't be used. Am I using it the wrong way? $\endgroup$ – William Riddle May 1 '15 at 15:27
  • $\begingroup$ After Mathematica gave me a nice result, I became curious as to the specific result of the limit of the terms for $p=2$ (and greater even $p$s). Since it probably is not particularly relevant to this question, I posted a new question $\endgroup$ – Peter Woolfitt May 1 '15 at 15:31

To me this is a simple application of Raabe-Duhamel's test. You'll get convergence for $p<1$ and, if $b-a>1$, you also get convergence for $p=1$. For $p>1$ the series diverges.

  • $\begingroup$ Thank you for letting me know this rule! It is really helpful to this problem. $\endgroup$ – William Riddle May 1 '15 at 16:36
  • $\begingroup$ @WilliamRiddle: The Raabe-Duhamel test is the natural thing to try when the root test (or the ratio test, which is often equivalent) fails. Convince yourself of its usefulness by studying $\sum \frac {(2n)!} {n!^2}$ first with the ratio, then with the Raabe-Duhamel test. $\endgroup$ – Alex M. May 1 '15 at 16:53

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