# How to find the sum of this infinite series: $\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$ [closed]

How do I find this particular sum? $$\sum_{n=1}^{ \infty } \frac1n \cdot \frac{H_{n+2}}{n+2}$$

where $H_n = \sum_{k=1}^{n}\frac1k$.

This was given to me by a friend and I have absolutely no idea how to proceed as I have never done these questions before. If possible, please give a way out without using polylogarithmic functions or other non-elementary functions.

• I dont think that this summation, if converges (it seems), can be expressed in elementary functions. – Masacroso Jan 19 '17 at 4:02
• The answer is $$\frac{1}{12}( 6+ \pi^2)$$ – S.C.B. Jan 19 '17 at 4:04
• If you use partial fractions, you can get it to partially telescope. – Michael Burr Jan 19 '17 at 4:05
• @S.C.B. Your answer is different from that given by renascence below :/ – Sawarnik Jan 19 '17 at 4:21
• @Dr.MV That was the answer I got from Wolframalpha. Apparently I mistyped something. Sorry. :/ – S.C.B. Jan 19 '17 at 5:35

Actually the calculation for this sum is very simple and what we need is the sum of telescopic series. In fact \begin{align} \sum_{n=1}^\infty\frac{H_{n+2}}{n(n+2)}&=\frac12\sum_{n=1}^\infty H_{n+2}\left(\frac{1}{n}-\frac1{n+2}\right)\\\\ &=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{1}{n}\left(H_{n}+\frac{1}{n+1}+\frac{1}{n+2}\right)-\frac{H_{n+2}}{n+2}\right)\\\\ &=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{H_n}{n}+\frac{1}{n(n+1)}+\frac{1}{n(n+2)}-\frac{H_{n+2}}{n+2}\right)\\\\ &=\frac{1}{2}\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac{H_{n+2}}{n+2}\right)+\frac12\sum_{n=1}^\infty\left(\frac{1}{n(n+1)}+\frac{1}{n(n+2)}\right)\\\\ &=\frac12\left(H_1+\frac{H_2}2\right)+\frac78\\\\ &=\frac{7}{4}. \end{align}

• (+1) Faster than mine! – Mark Viola Jan 19 '17 at 16:52
• @xpaul: Elegant and instructive! (+1) – Markus Scheuer Jan 19 '17 at 21:48
• +1 I believe there should not be a simpler solution compared to this. – Paramanand Singh Jan 20 '17 at 5:33

I thought it would be instructive to present a way forward that relies on elementary analysis, including knowledge of the sum $\sum_{k=1}^\infty \frac1{k^2}=\pi^2/6$, partial fraction expansion, and telescoping series. It is to that end we proceed.

The Harmonic number, $H_{n+2}$, can be written as

$$H_{n+2}=\sum_{k=1}^{n+2}\frac1k=\frac{1}{n+2}+\frac{1}{n+1}+\sum_{k=1}^n \frac1k$$

Therefore, we have

\begin{align} \sum_{n=1}^\infty\frac{1}{n(n+2)}\sum_{k=1}^{n+2}\frac1k&=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty \frac1k \sum_{n=k}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+2)}\right)\\\\ &=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\sum_{k=1}^\infty\frac1{2k^2}+\frac12\\\\ &=\sum_{n=1}^\infty\frac{1}{n(n+2)^2}+\sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)}+\frac{\pi^2}{12}+\frac12\\\\ &=\color{red}{\sum_{n=1}^\infty\left(\frac{1}{4n}-\frac{1}{4(n+1)}-\frac{1}{2(n+2)^2}\right)}\\\\ &+\color{blue}{\sum_{n=1}^\infty\left(\frac{1}{2n}-\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{2(n+1)}\right)}\\\\ &+\frac{\pi^2}{12}+\frac12\\\\ &=\color{blue}{1-\frac{\pi^2}{12}}+\color{red}{\frac14}+\frac{\pi^2}{12}+\frac12\\\\ &=\frac74 \end{align}

• How did you get the last term in first line of your derivation? It should be $\sum H_n/n(n+2)$. – Paramanand Singh Jan 19 '17 at 4:56
• I have to say I like this one a lot better than the other. "Heavy artillery to kill a flea" indeed. – user361424 Jan 19 '17 at 5:01
• @ParamanandSingh Nice catch! I've edited. Thank you my friend! And Happy New Year. -Mark – Mark Viola Jan 19 '17 at 5:01
• @user361424 Thank you!! Much appreciative. And Happy New Year. -Mark – Mark Viola Jan 19 '17 at 5:01
• @Sawarnik: it's not necessary to introduce $\pi$. Also the desired transformation from line 1 to line 2 of the derivation is not that difficult. We have $$\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k}\sum_{n=k}^{\infty}\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{2}\sum_{k=1}^{\infty}\frac{1}{k}\left(\frac{1}{k}+\frac{1}{k+1}\right)$$ and this leads to $$\frac{1}{2}\sum\frac{1}{k^{2}}+\frac{1}{2}\sum\frac{1}{k(k+1)}$$ The second sum easily evaluates to $1$ via telescoping. – Paramanand Singh Jan 19 '17 at 10:11

recall: $\displaystyle H_a = \int_0^1 \frac{1-x^a}{1-x}\,\mathrm{d}x$, and integration by parts once we have $$\int_0^1 x^{a-1} \ln (1-x)\,\mathrm{d}x = -\frac{H_a}{a}$$ Thus, \begin{align*}&\sum\limits_{n=1}^{\infty} \frac{H_{n+a}}{n(n+a)} = -\sum\limits_{n=1}^{\infty} \frac{1}{n}\int_0^{1} x^{n+a-1} \ln (1-x)\,\mathrm{d}x= \int_0^{1} x^{a-1} \ln^{2}(1-x)\,\mathrm{d}x\\&= \lim\limits_{b \to 1}\frac{\partial^2}{\partial b^2}\int_0^{1} x^{a-1}(1-x)^{b-1}\,\mathrm{d}x= \lim\limits_{b \to 1} \frac{\partial^2}{\partial b^2} \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\\&= \lim\limits_{b \to 1}\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\left((\psi_0(b) - \psi_0(a+b))^2 + \psi_1(b) - \psi_1(a+b)\right)\\&= \frac{1}{a}\left((\gamma + \psi_0(a+1))^2 + \frac{\pi^2}{6} - \psi_1(a+1)\right)\end{align*} Hence put $a=2$ we get $$\sum\limits_{n=1}^{\infty} \frac{H_{n+2}}{n(n+2)} =\frac{1}{2}\left [ \left ( \gamma +\underset{\psi _{0}\left ( 3 \right )}{\underbrace{\frac{3}{2}-\gamma}} \right )^{2}+\frac{\pi ^{2}}{6}-\underset{\psi _{1}\left ( 3 \right )}{\underbrace{\left (\frac{\pi ^{2}}{6}-\frac{5}{4} \right )} }\right ]=\frac{7}{4}$$

• Nice and inspiring solution, for sure. – Claude Leibovici Jan 19 '17 at 5:41

This is not an answer but it is too long for comment.

Just out of curiosity and playing with a CAS, $$S_p=\sum_{n=1}^{ p } \frac{H_{n+2}}{n(n+2)}=\frac{-2 \left(2 p^2+9 p+9\right) H_{p+3}+(7 p^3+38 p^2+64 p+33)}{4 (p+1) (p+2) (p+3)}$$ Using the asymptotics of harmonic numbers $$S_p=\frac{7}{4}-\frac{\log \left(p\right)+\gamma +1}{p}+O\left(\frac{1}{p^2}\right)$$

Similarly, considering $$T_p^{(a)}=\sum_{n=1}^{ p } \frac{H_{n+a}}{n(n+a)}$$ we get similar forms $$T_p^{(1)}=\frac{-(p+2) H_{p+2}+(2 p^2+5 p+3)}{(p+1) (p+2)}$$

$$T_p^{(3)}=\frac{-18 \left(3 p^3+24 p^2+59 p+44\right) H_{p+4}+(85 p^4+796 p^3+2624 p^2+3563 p+1650)}{54 (p+1) (p+2) (p+3) (p+4)}$$ for which the asymptotics are $$T_p^{(1)}=2-\frac{\log \left(p\right)+\gamma +1}{p}+O\left(\frac{1}{p^2}\right)$$ $$T_p^{(2)}=\frac{7}{4}-\frac{\log \left(p\right)+\gamma +1}{p}+O\left(\frac{1}{p^2}\right)$$ $$T_p^{(3)}=\frac{85}{54}-\frac{\log \left(p\right)+\gamma +1}{p}+O\left(\frac{1}{p^2}\right)$$ $$T_p^{(4)}=\frac{415}{288}-\frac{\log \left(p\right)+\gamma +1}{p}+O\left(\frac{1}{p^2}\right)$$ $$T_p^{(5)}=\frac{12019}{9000}-\frac{\log \left(p\right)+\gamma +1}{p}+O\left(\frac{1}{p^2}\right)$$ For sure, all limits correspond to Renascence_5's answer which can also write $$\frac 1a\left(\frac{\pi^2}6+\left(H_a\right){}^2-\psi ^{(1)}(a+1)\right)$$

• Very useful! (+1) – Mark Viola Jan 19 '17 at 16:14

There is an entirely elementary approach to the computation, using only partial fractions and the integral test.

Consider $$\sum_{n=1}^\infty\frac{1}{n}\cdot\frac{H_{n+2}}{n+2}$$ where $H_{n+2}$ is the $(n+2)$-th partial sum of the harmonic series. Consider the $k$-th partial sum of the given series: $$\sum_{n=1}^k\frac{1}{n}\cdot\frac{H_{n+2}}{n+2}.$$ Using partial fraction decomposition, we observe that $$\frac{1}{n}\cdot\frac{1}{n+2}=\frac{1}{2n}-\frac{1}{2(n+2)}$$ Therefore, the partial sum can be rewritten as $$\sum_{n=1}^k\left(\frac{H_{n+2}}{2n}-\frac{H_{n+2}}{2(n+2)}\right)=\sum_{n=1}^k\frac{H_{n+2}}{2n}-\sum_{n=1}^k\frac{H_{n+2}}{2(n+2)}$$ Using the change of variables $m=n+2$ on the second sum, we get $$\sum_{n=1}^k\left(\frac{H_{n+2}}{2n}-\frac{H_{n+2}}{2(n+2)}\right)=\sum_{n=1}^k\frac{H_{n+2}}{2n}-\sum_{m=3}^{k+2}\frac{H_m}{2m}.$$ We can observe that these sums partially telescope; in fact, the expression equals $$\frac{H_3}{2}+\frac{H_4}{4}+\sum_{n=3}^k\frac{H_{n+2}-H_n}{2n}-\frac{H_{k+1}}{2(k+1)}-\frac{H_{k+2}}{2(k+2)}.$$ Let's deal with each part of the expression separately.

Part 1: The first two terms can be evaluated directly. Namely, \begin{align*} H_3&=\frac{11}{6}&H_4&=\frac{25}{12}. \end{align*} Therefore, $$\frac{H_3}{2}+\frac{H_4}{4}=\frac{23}{16}.$$

Part 3: Using the integral test, we can bound the value of $H_{k+1}$ and $H_{k+2}$. Recall that for partial sums, the integral test's proof tells us that $$\int_2^{k+1}\frac{1}{x}dx\leq \sum_{n=1}^k\frac{1}{n}\leq\int_1^k\frac{1}{x}dx.$$ Therefore, we know that $$\ln(k+1)-\ln(2)\leq H_k\leq \ln(k).$$ Applying these inequalities in our case, we have to change $k$ to $k+1$ and $k+2$. Either way, however, we get $$\frac{\ln(k+2)-\ln(2)}{2(k+1)}+\frac{\ln(k+3)-\ln(2)}{2(k+2)}\leq\frac{H_{k+1}}{2(k+1)}+\frac{H_{k+2}}{2(k+2)}\leq\frac{\ln(k+1)}{2(k+1)}+\frac{\ln(k+2)}{2(k+2)}$$ The fractions on the left and right both go to zero as $k$ increases because they are logarithms over linear terms and linear terms grow faster. Use l'Hopital's rule once to check, if you need further justification. Therefore, these terms can be ignored in the final sum.

Part 2: We are now considering the sum $$\sum_{n=3}^k\frac{H_{n+2}-H_n}{2n}.$$ Observe that $H_{n+2}-H_n=\frac{1}{n+1}+\frac{1}{n+2}$. Therefore, we must compute $$\sum_{n=3}^k\left(\frac{1}{n+1}\cdot\frac{1}{2n}+\frac{1}{n+2}\cdot\frac{1}{2n}\right).$$ Once again, we use partial fraction decomposition. Observe that \begin{align*} \frac{1}{n+1}\cdot\frac{1}{2n}&=\frac{1}{2n}-\frac{1}{2(n+1)}& \frac{1}{n+2}\cdot\frac{1}{2n}&=\frac{1}{4n}-\frac{1}{4(n+2)}. \end{align*} This partial sum then becomes $$\sum_{n=3}^k\left(\frac{1}{2n}-\frac{1}{2(n+1)}+\frac{1}{4n}-\frac{1}{4(n+2)}\right)=\sum_{n=3}^k\frac{1}{2n}-\sum_{n=3}^k\frac{1}{2(n+1)}+\sum_{n=3}^k\frac{1}{4n}-\sum_{n=3}^k\frac{1}{4(n+2)}.$$ Using the change of variables $m=n+1$ in the second sum and $m=n+2$ in the fourth sum, we get $$\sum_{n=3}^k\frac{1}{2n}-\sum_{m=4}^{k+1}\frac{1}{2m}+\sum_{n=3}^k\frac{1}{4n}-\sum_{m=5}^{k+2}\frac{1}{4m}.$$ Once again, these terms telescope to $$\frac{1}{6}-\frac{1}{2(k+1)}+\frac{1}{12}+\frac{1}{16}-\frac{1}{4(k+1)}-\frac{1}{4(k+2)}.$$ All of the terms with $k$ in the denominator will vanish as $k$ increases, leaving $$\frac{1}{6}+\frac{1}{12}+\frac{1}{16}=\frac{5}{16}.$$

Adding everything up, we get $$\frac{23}{16}+\frac{5}{16}+0=\frac{7}{4}.$$

• Since $H_{n} = \sum (1/n)$ we can see from Cesaro-Stolz that $H_{n} / n \to 0$ as $n \to \infty$. This seems easier than approximating $H_{n}$ by $\log n$. But +1 for the simplest answer so far. – Paramanand Singh Jan 19 '17 at 12:06