I wanted to test the convergence of $1/n^{ln3}$ as n tends to infinity. I applied the integral test and found that the integral is (ln n) which certainly diverges as n tends to infinity. But the answer given is that this given series is converging.Would anybody help? Please.

  • $\begingroup$ So the integral test gives $\int_1^\infty \frac{1}{x^{\log(3)}}\,dx=\frac{1}{\log(3)-1}$ $\endgroup$ – Mark Viola May 14 '17 at 4:37
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    $\begingroup$ The integral of $1/n^{\log 3}$ is certainly not $\log n$. $\endgroup$ – mjqxxxx May 14 '17 at 4:38

This is a p-series and $\log(3) > 1$, hence it converges. You can use the comparison test to show this.


You can use the Cauchy Condensation Test, or the Integral Comparison Test, to show that $\sum_n n^{-p}$ converges iff $p>1.$

Let $(A(n))_n$ be a sequence that is monotonic except for finitely many $n $. Then:

  1. (Cauchy). $\sum_nA(n)$ converges iff $\sum_n 2^n A(2^n)$ converges.

  2. (Integral). If $f:\mathbb R^+\to \mathbb R$ is monotonic and continuous, and if $f(n)=A(n)$ for all but finitely many $n$, then $\sum_nA(n)$ converges iff $\int_1^xf(t)dt$ converges as $x\to \infty.$

By repeated use of the Cauchy Condensation, you can show that each of $\sum_n1/(n (\ln n)^p), \;\sum_n 1/(n (\ln n)(\ln \ln n)^p),...$ converges iff $p>1.$

The idea behind the Cauchy Test is illustrated by the cases $\sum_n 1/n$ and $\sum_n 1/n^2:$

We have $$\sum_{n=1}^{\infty}1/n=1+1/2 +(1/3+1/4)+(1/5+...+1/8)+(1/9+...+1/16)+...>$$ $$>1+2(1/2)+2^1(1/4)+2^2(1/8)+2^3(1/16)+...=$$ $$=1+1/2+1/2+1/2+....=\infty.$$

We have $$\sum_{n=1}^{\infty}1/n^2=1+(1/2^2+1/3^2)+(1/4^2...+1/7^2)+(1/8^2+...1/15^2)+...<$$ $$<1+2^1(1/2^2)+2^2(1/4^2)+2^3(1/8^2)=$$ $$=1+2^{-1}+2^{-2}+2^{-3}+...=2.$$


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