# Convergence of harmonic series multiplied by partial sums of harmonic series

How can I show if the following series is convergent or divergent $$\sum_{n=1}^\infty \frac1n \left ( 1 + \frac12 + \frac13 + \cdots + \frac1n \right) = \sum_{n = 1}^\infty\frac{1}{n}\sum_{k = 1}^n\frac{1}{k}$$ If it were a sequence I could have easily used Cauchy first theorem.

• This isn't a series, it is a sequence. Why do you call it a series in the first sentence and claim it is not a sequence in the second? Do you mean $\sum_{n=1}^\infty$ of the sequence you've written? Please clarify your question. – Erick Wong Jan 1 '17 at 6:11
• It is a series once you replace n by k and put an summation in front of it where k goes form – Nitish Jan 1 '17 at 6:13
• From 1 to n.... – Nitish Jan 1 '17 at 6:14
• @Nitish Please verify that ErickWong's edit is what you intended. – Patrick Stevens Jan 1 '17 at 6:15
• "Yes. That's what I was asking" What is "that"? Sorry but the series $\sum\frac1n ( 1 + \frac12 + \frac13 + \cdots + \frac1n )$ is (obviously) divergent and the sequence $\frac1n ( 1 + \frac12 + \frac13 + \cdots + \frac1n )$ is (almost as obviously) convergent. Reading the comments, my impression is that the OP does not know what they want to ask. – Did Jan 1 '17 at 11:06

Comparison: $$\sum_{n=1}^\infty \frac1n \left( 1 + \frac12 + \frac13 + \cdots + \frac1n \right) \ge \sum_{n=1}^\infty \frac 1 n = \infty.$$

Admitting that you can use harmonic number, let us consider $$S_p= \sum_{n = 1}^p\frac{1}{n}\sum_{k = 1}^n\frac{1}{k}=\sum_{n = 1}^p\frac{H_n}{n}=\frac{\left(H_p\right){}^2}{2}-\frac{H_p^{(2)}}{2}-\psi ^{(1)}(p+1)+\frac{\pi ^2}{6}$$ Considering the asymptotics $$H_p=\gamma +\log \left(p\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^4}\right)$$ $$H_p^{(2)}=\frac{\pi ^2}{6}-\frac{1}{p}+\frac{1}{2 p^2}-\frac{1}{6 p^3}+O\left(\frac{1}{p^4}\right)$$ $$\psi ^{(1)}(p+1)=\frac{1}{p}-\frac{1}{2 p^2}+\frac{1}{6 p^3}+O\left(\frac{1}{p^4}\right)$$ you should end with $$S_p=\frac{\pi ^2}{12}+\frac{\gamma ^2}{2}+\gamma \log \left(p\right)+\frac{1}{2} \log ^2\left(p\right)+\frac{\log \left({p}\right)+\gamma -1}{2 p}+O\left(\frac{1}{p^2}\right)$$ which shows the result.

Computing exactly for $p=10$, we have $$S_{10}=\frac{32160403}{6350400}\approx 5.06431$$ while the above expression gives $\approx 5.06308$.

• I'm still an undergraduate....I'm not qualified enough to learn this...haha – Nitish Jan 1 '17 at 8:03
• @Nitish. So, just forget it for the time being. In a few years, look at it again. By the way Happy New Year and good luck in your studies. – Claude Leibovici Jan 1 '17 at 8:21
• Haha, I went back to some of my old posts. Very old, and it sure is a shocker to see how much one changes in one years worth of time. – Simply Beautiful Art Jan 1 '17 at 12:58

The series diverges. You have that $$1+1/2+\cdots \geq 1$$ and thus you only have to look at $$\sum_{n=1}^\infty n^{-1}$$ which is a lower bound to your series. The latter series is the harmonic series that is well-known to diverge.

EDIT: I answered the question as it was originally stated, but I think you intended to ask a different question from the one you actually asked.

This is the mean of $S_n = \{1, \frac{1}{2}, \dots, \frac{1}{n}\}$. That's clearly bounded by $0$ and $1$, and also is clearly decreasing because to move from $S_n$ to $S_{n+1}$ we don't remove anything from $S_n$ but we add a smaller thing (namely $\frac{1}{n+1}$).