# convergence of infinite series $\sum\limits_{n=0}^{\infty}\frac{(1+1/2+1/3+…+1/n)}{n}$

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 .

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

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 $$\infty$$.

• yes Thanks , I failed to see this I will edit the question – kira0705 Oct 2 '18 at 13:42
• @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) – Eff Oct 2 '18 at 13:44
• okay , should I add under edit that the series should be divergent , without changing the question entirely ? – kira0705 Oct 2 '18 at 13:45
• @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. – Eff Oct 2 '18 at 13:49

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.

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

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.

• 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. – TheSilverDoe Oct 2 '18 at 14:34
• Can you point out my mistake? I am unable to find it. – Avinash N Oct 2 '18 at 14:55
• 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. – TheSilverDoe Oct 2 '18 at 14:59
• 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. – Avinash N Oct 2 '18 at 15:29
• 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. – TheSilverDoe Oct 2 '18 at 15:39

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 $$N\to +\infty$$.