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The series is: $\displaystyle\sum_{n=1}^\infty \dfrac1{n (\ln n)^p}$

I don't know what to do from here since $p$ is on $\ln$.

Would $p$ still have to be $> 1$ since $\ln$ is changing in terms of $n$?

user
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user3201346
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3 Answers3

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The Cauchy condensation test easily gives that your series converges iff $p>1$.

Jack D'Aurizio
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  • I have a little question, note that $\sum 2^{n}a_{2^{n}}=\sum \frac{1}{2^n(ln(2^n))^p}$, but this is equal to $\sum \frac{1}{n^{p}}$ but why from this $(ln(2^n))^p$ to this $n^{p}$? – Wrloord Aug 27 '23 at 23:49
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$\textbf{Hint:}$ Use Integral test for convergence.

By Integral test of convergence you calculate the integral $\displaystyle \int_{1}^{\infty} \frac{1}{x(\ln x)^p}dx=\int_{0}^{\infty} \frac{1}{t^p} dt$ (substitution $t=\ln x$).

For $p=-1$ you have $\displaystyle \int \frac{1}{t^p} dt=\log|t|+C$, so integral diverges.

For $p \neq -1 $ you have $\displaystyle \int \frac{1}{t^p} dt=\frac{p-1}{t^{p-1}}+C$, so for $p<-1$ integral diverges, for $p>-1$ converges.

agha
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  • This is clearly not true. If $p = -\frac{1}{2}$ then $\frac{1}{n(\ln{n})^p} = \frac{\ln{n}^{\frac{1}{2}}}{n} > \frac{1}{n}$ (for $n > 3$) so the series must diverge. – Davis Yoshida Aug 17 '14 at 22:53
  • That should say "for $p < 1$ the integral diverges, for $p > 1$ converges". – JimmyK4542 Aug 17 '14 at 23:00
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Hint: By the integral test, $\displaystyle\sum_{n=1}^\infty \dfrac1{n (\ln n)^p}$ converges iff $\displaystyle\int_{1}^{\infty}\dfrac{dx}{x(\ln x)^p}$ converges.

Now substitute $u = \ln x$ into that integral, and see what you get.

JimmyK4542
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