3
$\begingroup$

I have applied Cauchy condensation test for to test the convergence the series $\sum_{n=2}^{\infty}\frac{1}{{(\log n)}^{p}}$, where p is constant, I got $\frac{1}{{\log 2}^{p}}\sum_{k=1}^{\infty}\frac{2^{k}}{k^{p}}$ . I do not understand for which value of p such that the original series is convergent. Also have used Cauchy integral test but did not solve the improper integral $\int_{2}^{\infty} {\frac{1}{(\log{x})^{p}}}dx$. I do not understand the convergent or not, if it convergent what is the value of p will be. Please some one help me. Thanks

$\endgroup$
6
$\begingroup$

Hint. Using the Cauchy condensation test, you are led to consider convergence/divergence of the series $$ \sum_{k=1}^{\infty}\frac{2^{k}}{k^{p}}. $$ But, for any fixed $p$, we have $$ \lim_{k\to\infty}\frac{2^{k}}{k^{p}} \neq 0 $$ thus your inital series is always divergent.

$\endgroup$
0
$\begingroup$

The integral test should work too. Setting $logx=t$ gives $dx=e^tdt$ gives a new integral of the form $\frac{e^t}{t^n}$ integrating to infinity. I see an e-power in the numerator against a polynomial term in the denominator so I am getting "wet feet" here. Can you finish it?

$\endgroup$

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.