0
$\begingroup$

I am trying to prove that below infinite series is convergent $$\sum\limits_{n=1}^{\infty}\frac{(1+1/2+1/3+...+1/n)}{n}$$

Edit : as pointed out the series is divergent and hence my approach is wrong

My approach :

I tried to approach it by Dirichlet convergence theorem by taking $u_{n}=\sum\limits_{n=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right) $ and $v_{n}=\frac{1}{n}$

but got struck at how to prove that the sum of $u_{n}$ upto n terms is bounded (if at all it is).

Earlier I had tried Ratio test through whichI was getting $\frac{u_{n}}{u_{n+1}}=1$ and thus failing the test .

Please let me know how can I approach this problem .

$\endgroup$
  • 4
    $\begingroup$ It doesn't look like convergent. Eact term is bounded below by $\frac{1}{n}$ and the harmonic series clearly diverges. $\endgroup$ – AdditIdent Oct 2 '18 at 13:29
  • 1
    $\begingroup$ Surely you didn't mean to start at $n=0$. $\endgroup$ – J.G. Oct 2 '18 at 13:44
  • $\begingroup$ Thanks for pointing it out I corrected it $\endgroup$ – kira0705 Oct 2 '18 at 13:49
6
$\begingroup$

Your series does not converge. Clearly, you have that $$\color{red}{1 + \frac12+...+\frac1n \geq 1}$$ for all $n\geq 1$. Hence, by comparison $$\sum\limits_{n=1}^N \frac{\color{red}{1+1/2+...+1/n}}{n}\geq \sum\limits_{n=1}^N \frac{\color{red}{1}}{n}$$ The right hand side diverges to $\infty$ as $N\to\infty$ and is known as the Harmonic series. Your series is strictly larger by comparison and hence also diverges to $\infty$.

$\endgroup$
  • $\begingroup$ yes Thanks , I failed to see this I will edit the question $\endgroup$ – kira0705 Oct 2 '18 at 13:42
  • $\begingroup$ @kira0705 If you made a mistake in your question, instead create a new question. (This site discourages completely changing the question after answers have been given) $\endgroup$ – Eff Oct 2 '18 at 13:44
  • $\begingroup$ okay , should I add under edit that the series should be divergent , without changing the question entirely ? $\endgroup$ – kira0705 Oct 2 '18 at 13:45
  • $\begingroup$ @kira0705 Yes, it's fine to change that series should be divergent. I'm simply saying if you wrote the completely wrong series or something like that, then rather ask in a new question. $\endgroup$ – Eff Oct 2 '18 at 13:49
5
$\begingroup$

The general term of your series is equivalent to $$\frac{\ln(n)}{n}$$

which is the general term of a divergent series. Therefore, your series diverges.

$\endgroup$
  • $\begingroup$ Thanks , I will make an edit to the question $\endgroup$ – kira0705 Oct 2 '18 at 13:42
0
$\begingroup$

Consider the partial sum of the given series. 1, 1/2, 11/18, 25/48...

By doing some calculation we have show that the sequence of partial sums form an non-increasing sequence and it is clearly bounded below (bounded also). Thus , this sequence of partial sums is convergent. So that the given series is convergent.

Edit: I have noticed that made a mistake. This series is diverges. Thanks to Thesilver doe.

$\endgroup$
  • $\begingroup$ Well, it's explicitely written in the post that the author made a mistake and the series is divergent... as explained in the other answers. $\endgroup$ – TheSilverDoe Oct 2 '18 at 14:34
  • $\begingroup$ Can you point out my mistake? I am unable to find it. $\endgroup$ – Avinash N Oct 2 '18 at 14:55
  • $\begingroup$ Well, I don't know which "calculation" you did, but of course the sequence of partial sums is non decreasing ! The terms of your series are all positive, so the sequence of partial is obviously increasing. $\endgroup$ – TheSilverDoe Oct 2 '18 at 14:59
  • $\begingroup$ I have got like this. Let S(n) be the n th partial sum of the given series. I have got that. S(n)-S(n+1)= 1/2 + 1/3 +...+1/(n+1) >0. Thus it is decreasing sequence. $\endgroup$ – Avinash N Oct 2 '18 at 15:29
  • 1
    $\begingroup$ You are making a confusion between the sequence of the general terms of the series (that is $\frac{1+\frac{1}{2} + ... \frac{1}{n}}{n}$, which is indeed decreasing) and the sequence of the partial sums of that series, which is increasing. $\endgroup$ – TheSilverDoe Oct 2 '18 at 15:39
0
$\begingroup$

By summation by parts

$$ \sum_{n=1}^{N}\frac{H_n}{n} = H_N^2 - \sum_{n=1}^{N-1}\frac{H_n}{n+1} $$ hence $$ -\frac{H_N}{N+1}+\sum_{n=1}^{N}\frac{H_n}{n} = H_N^2-\sum_{n=1}^{N}\frac{H_{n+1}-\frac{1}{n+1}}{n+1}$$ and $$ -\frac{H_N}{N+1}+\sum_{n=1}^{N}\frac{H_n}{n}+\sum_{n=2}^{N+1}\frac{H_n}{n}=H_N^2+\sum_{n=2}^{N+1}\frac{1}{n^2}, $$

$$ 2\sum_{n=1}^{N}\frac{H_n}{n}=H_N^2+\sum_{n=1}^{N}\frac{1}{n^2}, $$ $$ \sum_{n=1}^{N}\frac{H_n}{n}=\frac{H_N^2+H_N^{(2)}}{2}=\frac{1}{2}\log^2 N+\gamma\log N+\frac{\gamma^2+\zeta(2)}{2}+O\left(\frac{\log N}{N}\right) $$ as $N\to +\infty$.

$\endgroup$

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.