Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose: $$\sum_{n=2}^{\infty} \left( \frac{1}{n(\ln(n))^{k}} \right) =\frac{1}{ 2(\ln(2))^{k} } +\frac{1}{ 3(\ln(3))^{k} }+..., $$ by which $k$ does it converge?

When I use comparison test I get inconclusive result:

$\lim_{n\rightarrow\infty} \frac{u_{n+1}}{u_{n}}=\frac{n\ln(n)^{k}}{(n+1)\ln(n+1)^{k}} =\lim_{n\rightarrow\infty} =\frac{n\ln(n)^{k}+\ln(n+1)^{k}-\ln(n+1)^{k}}{(n+1)\ln(n+1)^{k}}\approx 1- \frac{\ln(n+1)^{k}}{(n+1)\ln(n+1)^{k}}=\\1-\lim_{n\rightarrow \infty}\frac{1}{n+1}=1$

Now my conclusion would be when $k\in\mathbb R$ but I feel I am doing something wrong because I am pretty sure I have done this kind of problems earlier where I used some well-known series comparison. WA nothing here.

[Update] Trying to use Cauchy condensation test

$$\sum_{n=2}^{\infty}\left( n^{1/k} ln(n)\right)^{-k}$$

and now C-test:

$$\sum_{n=2}^{\infty}\left( \left(2^{n} \right) 2^{n/k} ln(2^{n})\right)^{-k}=$$ $$\sum_{n=2}^{\infty} e^{-k \left( n(1+\frac{1}{k})ln(2)+ln(n)+ln(ln(2) \right) }$$

so now as a geometric series, can I conclude something in terms of $k$? Look $k$ is still in one denominator not the just first factor in the exponent.

share|cite|improve this question
One uses the Integral Test (or Cauchy Condensation). By integrating, you can show that the integral of $1/(x\,\log^k x)$ from say $2$ to infinity converges if $k>1$ and diverges if $k\le 1$. – André Nicolas Feb 11 '12 at 7:14
@AndréNicolas: am I doing it totally wrong way (see my updated question with example)? I cannot understand what the word integral means here when I look at the wikipedia here. – hhh Feb 11 '12 at 16:03
You misunderstood my comment (my fault, too condensed). I meant that you can use either Integral Test or Cauchy Condensation. Integral Test is probably taught in all courses that deal with convergence of series. I mentioned Cauchy Condensation as an alternative. It is taught much less often. – André Nicolas Feb 11 '12 at 16:30
up vote 2 down vote accepted

For completeness, we sketch the Integral Test approach.

Let $f$ be a function which is defined, non-negative, and decreasing (or at least non-increasing from some point $a$ on. Then $\sum_1^\infty f(n)$ converges if and only if the integral $\int_a^\infty f(x)\,dx$ converges.

In our example, we use $f(x)=\dfrac{1}{x\,\ln^k(x)}$. Note that $f(x)\ge 0$ after a while, and decreasing So we want to find the values of $k$ for which $$\int_2^\infty \frac{dx}{x\ln^k(x)}\qquad\qquad(\ast)$$ converges (we could replace $2$ by say $47$ if $f(x)$ misbehaved early on).

Let $I(M)=\int_0^M f(x)\,dx$. We want to find the values of $k$ for which $\lim_{M\to\infty} I(M)$ exists.

It is likely that this was already done when you were studying improper integrals, but we might as well do it again.

Suppose first that $k>1$. To evaluate the integral, make the substitution $\ln x=u$. Then $$I(M)=\int_2^M \frac{dx}{x\ln^k(x)}=\int_{\ln 2}^{\ln M} \frac{du}{u^k}.$$ We find that $$I(M)=\frac{1}{k-1}\left(\frac{1}{(\ln 2)^{k-1}}- \frac{1}{(\ln M)^{k-1}}\right).$$ Because $k-1>0$, the term in $M$ approaches $0$ as $M\to\infty$, so the integral $(\ast)$ converges.

By the Integral Test, we therefore conclude that our original series converges if $k>1$.

For $k=1$, after the $u$-substitution, we end up wanting $\int\frac{du}{u}$. We find that $$I(M)=\ln(\ln M)-\ln(\ln 2).$$ As $M\to\infty$, $\ln(\ln M)\to \infty$ (though glacially slowly). So by the Integral Test, our series diverges if $k=1$.

For $k<1$, we could again do the integration. But an easy Comparison with the case $k=1$ proves divergence.

share|cite|improve this answer

I think Cauchy's Condensation test will do the deal.

$$\sum a_n \text{converges} \iff \sum2^na_{2^n} \text{converges}$$

And, on applying this test, and simplifying the test, you'll need to compare it with the series $\sum n^{-p}$. For what values of $p$ does this series converge?

share|cite|improve this answer
...solved with this method +1, look at the page 2 of this paper here. Thank you for the key term. The proof requires that you use two-times Cauchy-condesation also for the term $\sum_{n=2}^{\infty}n^{-k}$ where you know that the geometric serie will convege when $k>1$. But what about the integral -method? – hhh Feb 11 '12 at 16:24
@hhh I am unaware of the details of the integral test. Probably, someone else might write up an answer if you really want to know a solution that uses integral test. – user21436 Feb 11 '12 at 16:41

You can use integral/series comparison.

share|cite|improve this answer
Something more elaborate would be appreciated... – J. M. Feb 11 '12 at 16:07
...err actually there must be other ways to solve this as well. What does it mean to compare to integral? – hhh Feb 11 '12 at 16:27
As the function $ x \longmapsto \frac{1}{x \ln(x)^{\alpha}}$ is decreasing, it is easy to check that the sum $\sum_{k=0}^n\frac{1}{k \ln(k)^{\alpha}}$ and the integral $\int_0^x\frac{dt}{t \ln(t)^{\alpha}}$ have the same behaviour as $n$ and $x$ go to infinity. – Selim Ghazouani Feb 11 '12 at 17:02
@SelimGhazouani Please add this kind of clarifications to your question to make it fleshy or something substantial. This answer is caught here at the Low Quality Posts which I agree with. I am downvoting because this is fundamentally shorter version of the first comment to the question. Write a substantial answer and ping me here. I'll retract it. – user21436 Feb 11 '12 at 17:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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