5
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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