# Proof that $\sum_{k=2}^{\infty} \frac{H_k}{k(k-1)}$ where $H_n$ is the sequence of harmonic numbers converges?

How to prove that $$\displaystyle \sum_{k=2}^{\infty} \dfrac{H_k}{k(k-1)}$$ where $$H_n$$ is the sequence of harmonic numbers converges and that $$\dfrac{H_n}{n(n-1)}\to 0 \$$

I have already proven by induction that this equals $$\left(2-\dfrac{1}{(n+1)}-\dfrac{h_{n+1}}{n} \right)$$ for every $$n\ge1$$ but am not sure how to use this in solving my problem. Could anyone give me some tips?

• @Peter What would be your preferred way to show that $\sum \ln (k) /(k(k-1))$ is convergent? The way I thought of is to show that $\frac{\log x}{x^2}$ is eventually decreasing by showing the derivative is eventually negative...then note that that integral is finite by integration by parts...then use the integral test (since monotone) against the sum $\frac{\log k}{k^2}$...then note that $k(k-1) \geq Ck^2$ for some $C$ eventually. But I'm interested to know if that is the way you would use. – Eric Auld Nov 27 '13 at 18:52
• Summing $H_k/(k(k-1)$ for $k$ from $2$ to $n=2$ there is only the one term $H_2/(2)=(3/2)/2=3/4.$ However your formula $2-1/(n+1)-(H_n+1)/n$ gives $5/12$ when $n=2$. Check and adjust the right formula, and then taking its limit will be easy because the term $H_n/n$ tends to $0$ as $n \to \infty.$ – coffeemath Nov 27 '13 at 18:52
• @coffeemath Sorry, I'm not really sure how to format on here but it isn't H_n +1 it is H_(n+1) , the n+1 is meant to be subscripted – user2659030 Nov 27 '13 at 19:00
• @Peter except that I have not learned what you said and I don't understand it – user2659030 Nov 27 '13 at 19:03
• @Peter I'm glad I asked, because your way is much faster and better. – Eric Auld Nov 27 '13 at 19:03

This just begs to be telescoped:

$$\sum_{k=2}^\infty\sum_{n=1}^k \frac{1}{k(k-1)n} = \sum_{k=2}^\infty \frac{1}{k(k-1)} + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty\frac{1}{k(k-1)}$$

$$= \sum_{k=2}^\infty (\frac{1}{k-1} -\frac{1}{k}) + \sum_{n=2}^\infty \frac{1}{n} \sum_{k=n}^\infty(\frac{1}{k-1} -\frac{1}{k})$$

$$= 1 + \sum_{n=2}^\infty \frac{1}{n}\cdot\frac{1}{n-1} =1 + \sum_{n=2}^\infty(\frac{1}{n-1}-\frac{1}{n}) =2$$

• Nice! Avoids any need for estimating harmonic series, etc. +1 – coffeemath Nov 27 '13 at 20:48

The closed form expression for the partial sum $S_n=\sum_{k=2}^n H_k/(k(k-1)$ is $$S_n=2-\frac{1}{n}-\frac{H_n}{n}.$$ This can be shown by induction, or more easily by writing out the sum and using that $1/(k(k-1)=1/(k-1)-1/k$ to collect the contributions of $1,1/2,1/3,\cdots,1/n$ to the sum. As others have pointed out, $H_n-\ln(n)\to C$ where $C$ is Euler's constant, and from that one gets $H_n/n \to 0$ in a few steps, maybe using L'Hopital on $\ln n / n.$

So the limit of the partial sum $S_n$ is $2$, which is also the sum from $2$ to infinity of the terms $H_k/(k(k-1))$. That the $k^{th}$ term goes to zero now follows since the series converges, or alternately it may be shown directly using the generous upper bound $H_n \le n$, so that $H_n/(n(n-1) \le 1/(n-1) \to 0.$

• Thank you for all your help. I hope this is more useful to others than for me since I have no mention of Euler's constant or that the harmonic series is bound by ln(n) in my text book. From what my book says, the partial sums of the harmonic series is unbounded. – user2659030 Nov 27 '13 at 19:55
• Yes, the partial sums of the harmonic series are unbounded, but in your problem these are being divided by $k(k-1)$ before summing them. I'll think a bit about how to approach this problem without Euler's constant or the estimate $h_k \approx ln(k).$ – coffeemath Nov 27 '13 at 20:00
• I think I have kind of figured it out myself – user2659030 Nov 27 '13 at 20:05