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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.

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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
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3 Answers

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.

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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?

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...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
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You can use integral/series comparison.

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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
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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
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