The sum of an infinite series containing a finite series in each denominator [duplicate]

Evaluate $$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\cdots+\frac{1}{\left[\frac{k(k+1)}{2}\right]^2}+\cdots$$ to $$\infty$$, where $$k$$ is the $$k$$th term of the series.

Using Microsoft Excel, I found that the sum of the first $$100$$ terms $$=1.15947\dots$$.

Does the convergence value have an exact form? If yes, what is it?

marked as duplicate by grand_chat, YuiTo Cheng, Nosrati, Lord Shark the Unknown, TheSimpliFireJun 22 at 8:05

• Your form is 4$\sum(1/(k^2+k)^2)$. Seems like that must be in some book somewhere... where did you get this from? – The Count Jun 22 at 0:49
• It converges according to my CAS, to $4·(\pi^2 - 9)/3$ – Sudix Jun 22 at 0:52
• WA gives the value of $4\sum \frac 1{(k(k+1))^2}=\frac 43\times (\pi^2-9)$. – lulu Jun 22 at 0:53
• @Sudix, I have checked that, it is true, THANKS!. But how could we approach that value? – Hussain-Alqatari Jun 22 at 0:56
• @Hussain-Alqatari It's the same... Just up to a multiple of 1/4. – YuiTo Cheng Jun 22 at 3:07

Using fractional decomposition we get:

$$\frac 1{\left(\frac{k(k+1)}{2}\right)^2} = \frac 4{(k + 1)^2} + \frac 8{(k + 1)} +\frac 4 {k^2} - \frac 8 k$$

So, your sum simplifies to one telescoping sum, and the sum $$\sum_{k=1}^\infty \frac 1 {k^2} =\frac{\pi^2} 6$$

So, all together we get:

$$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\dots+\frac{1}{[\frac{k(k+1)}{2}]^2}+\dots\\ = \sum_{k=1}^\infty \frac 1{\left(\frac{k(k+1)}{2}\right)^2} \\= \sum_{k=1}^\infty \frac 4{(k + 1)^2} + \frac 8{(k + 1)} +\frac 4 {k^2} - \frac 8 k \\ = 4\left( \sum_{k=1}^\infty \frac 1{(k + 1)^2} \right) + 4\left( \sum_{k=1}^\infty \frac 1 {k^2} \right)+ 8\left( \sum_{k=1}^\infty \frac 1{(k + 1)}- \frac 1 k \right)$$ In the first sum, you do an index shift to the left and add the first summand (as a zero). $$=4\left(-1+ \sum_{k=1}^\infty \frac 1{k^2} \right) + 4\left( \sum_{k=1}^\infty \frac 1 {k^2} \right)+ 8\left( \sum_{k=1}^\infty \frac 1{(k + 1)}- \frac 1 k \right) \\= -4+4\cdot \frac{\pi^2} 6 + 4\cdot \frac{\pi^2} 6 -8 = 4·(\pi^2 - 9)/3$$

• Can someone explain why these were downvoted? – dxdydz Jun 22 at 1:09
• I was wondering the same thing. But I do note that Sudix repeatedly writes $\left({k(k+1)\over2}\right)^2$ where what's meant is $\left({k(k+1)\over2}\right)^{-2}$. Maybe someone doesn't like typos? – Gerry Myerson Jun 22 at 1:12
• @GerryMyerson Well spotted! I did a broken copy & paste there... – Sudix Jun 22 at 1:16

As showed in answers, the infinit sum is not to hard. We could also have a good approximation of the partial sum $$S_p=\sum_{k=1}^p\frac{4}{k^2 (k+1)^2}$$ Using a CAS, $$S_p=\frac{4 \left((p+1)^2 \left(\pi ^2-6 \psi ^{(1)}(p+1)\right)-3 p (3 p+4)\right)}{3 (p+1)^2}$$ and using the asymptotics of the polygamma function $$S_p=\left(\frac{4 \pi ^2}{3}-12\right)-\frac{4}{3 p^3}+\frac{4}{p^4}+O\left(\frac{1}{p^5}\right)$$

For $$p=10$$, the exact value is $$\frac{22254209}{19209960}\approx 1.15847$$ while the above approximation gives $$\frac{4 \pi ^2}{3}-\frac{90007}{7500}\approx 1.15854$$.

$$\frac {1}{k(k+1)}= \frac {1}{k}-\frac {1}{k+1}$$

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

$$\sum_1^{\infty}\frac {1}{k^2(k+1)^2}= \sum_1^{\infty}\frac {1}{k^2}+\sum_1^{\infty}\frac {1}{(k+1)^2} -\sum_1^{\infty} \frac {2}{k(k+1)}=$$

$$\frac {\pi ^2}{6} +(\frac {\pi ^2}{6}-1)-2$$

$$\sum_1^{\infty} \frac {4}{k^2(k+1)^2} = 4(\frac {\pi^2 }{6} + \frac {\pi^2 }{6}-1 -2) = 4(\frac {\pi^2-9}{3})\approx 1.159472535...$$