# Evaluate the sum $\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}$?

Evaluate the series $$\sum_{k=0}^{\infty}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}=?$$

Can you help me ? This is a past contest problem.

-
You could try breaking it into partial fractions. – Jair Taylor Oct 30 '12 at 5:01
@JairTaylor: Or using by the Beta function, which gives a continuous analogue to partial fractions. – Eric Naslund Oct 30 '12 at 5:05
@Jair Taylor:Yes! in fact i know the results of this problem. I want to know some different method. Thank you very much. – Dao yi Peng Oct 30 '12 at 5:06
@EricNaslund:Thank you! I and you use the same method. I have used Gamma fuuction. – Dao yi Peng Oct 30 '12 at 5:12
Dao: Since you know the results, why not mention it (and where you read them)? – Did Oct 30 '12 at 7:14

## 4 Answers

This is Problem 2 of Day 1 of the 2010 International Mathematics Competition. If you go to the competition website, you can find several solutions, but here is a solution I came up with using the Beta function:

We have the identity ** $$\frac{3!}{(4l+1)(4l+2)(4l+3)(4l+4)}=\int_{0}^{1}x^{4l}(1-x)^{3}\text{d}x.$$ Thus we may write $$\sum_{l=0}^{\infty}\frac{1}{(4l+1)(4l+2)(4l+3)(4l+4)}=\sum_{l=0}^{\infty}\frac{1}{3!}\int_{0}^{1}x^{4l}(1-x)^{3}\text{d}x.$$ Rearranging the order of summation and integration yields $$\frac{1}{6}\int_{0}^{1}\left(\sum_{l=0}^{\infty}x^{4l}\right)(1-x)^{3}\text{d}x=\frac{1}{6}\int_{0}^{1}\frac{(1-x)^{3}}{1-x^{4}}\text{d}x=\frac{1}{6}\int_{0}^{1}\frac{(1-x)^{2}}{(1+x)(1+x^{2})}\text{d}x.$$ Using partial fractions to split up the integrand we then have $$\frac{1}{6}\int_{0}^{1}\frac{2}{1+x}-\frac{1+x}{1+x^{2}}\text{d}x=\frac{1}{3}\int_{0}^{1}\frac{1}{1+x}-\frac{1}{6}\int_{0}^{1}\frac{1}{1+x^{2}}\text{d}x-\frac{1}{6}\int_{0}^{1}\frac{x}{1+x^{2}}\text{d}x$$ and evaluating the integrals yields $$=\frac{1}{3}\log2-\frac{1}{6}\frac{\pi}{4}-\frac{1}{12}\log2=\frac{1}{4}\log2-\frac{\pi}{24}$$

** This identity follows from the fact that $\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ where $\text{B}(x,y)$ is the Beta Function, but it also follows directly from expanding and integrating since $l$ is an integer. Essentially, this integral encodes the partial fraction decomposition.

-
Where did the 3! come from? – Jair Taylor Oct 30 '12 at 5:21
@JairTaylor: It is a $\Gamma(4)$ because there are $4$ terms. – Eric Naslund Oct 30 '12 at 6:26
Sledgehammers get the job done, too! :-) – robjohn Oct 31 '12 at 19:03

Let \begin{align}S_n & =\sum_{k=0}^{n}\frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}\\ & =\sum_{k=0}^{n}\left(\frac{1}{6}\frac{1}{4k+1}-\frac{1}{2}\frac{1}{4k+2}+\frac{1}{2}\frac{1}{4k+3}-\frac{1}{6}\frac{1}{4k+4}\right). \end{align} $$A_n=\sum_{k=0}^{n}\left(\frac{1}{4k+1}-\frac{1}{4k+3}\right),$$ $$B_n=\sum_{k=0}^{n}\left(\frac{1}{4k+1}-\frac{1}{4k+2}+\frac{1}{4k+3}-\frac{1}{4k+4}\right),$$ and $$C_n=\sum_{k=0}^{n}\left(\frac{1}{4k+2}-\frac{1}{4k+4}\right).$$ It is easy check that $$S_n=\frac{1}{3}B_n-\frac{1}{6}A_n-\frac{1}{6}C_n.$$ Therefore, $$\lim_{n\to\infty}S_n=\lim_{n\to\infty}\frac{2B_n-A_n-C_n}{6}=\frac{2\ln 2-\dfrac{\pi}{4}-\dfrac{1}{2}\ln 2}{6}=\frac{1}{4}\ln 2-\frac{\pi}{24}.$$

-
-1: This is an incorrect way though you end up getting the right answer. You cannot rearrange terms since after splitting it as partial fractions the sum diverges absolutely. – user17762 Oct 30 '12 at 15:51
This is the same reason why the method in my edit section ends up giving an incorrect answer. – user17762 Oct 30 '12 at 15:52
@Marvis: The decomposition Dao Yi Peng did can be justified, but undoubtedly he needs to say more than "It is easy to check that"... DaoYiPeng: In general if it is so easy, it should take less than half a sentence to explain why. If it takes more than that, then you should probably explain it anyway. – Eric Naslund Oct 30 '12 at 17:52
@EricNaslund True. He needs to argue out why the decomposition can be done in the first place. And actually I am interested in the justification since those series are not absolutely convergent. How/Why does his rearrangement give the right answer? – user17762 Oct 30 '12 at 17:53
There is still something wrong here. $A_n\to\frac\pi4$, $B_n\to\log(2)$, and $C_n\to\frac12\log(2)$. So as computed above, $S_n\to\frac\pi{12}-\frac14\log(2)$. I think that $S_n=\frac13B_n-\frac16A_n-\frac16C_n$ is correct – robjohn Oct 31 '12 at 19:37

\begin{align} f(k) & = \dfrac1{(4k+1)(4k+2)(4k+3)(4k+4)}\\ & = - \dfrac12 \dfrac1{4k+2} + \dfrac16 \dfrac1{4k+1} + \dfrac12 \dfrac1{4k+3} - \dfrac1{6} \dfrac1{4k+4} \end{align} $$f(k) = -\dfrac12 \int_0^1 t^{4k+1} dt + \dfrac16 \int_0^1 t^{4k} dt + \dfrac1{2} \int_0^1 t^{4k+2} dt - \dfrac1{6} \int_0^1 t^{4k+3} dt$$ $$S = \sum_{k=0}^{\infty} f(k) = \sum_{k=0}^{\infty} \left(-\dfrac12 \int_0^1 t^{4k+1} dt + \dfrac16 \int_0^1 t^{4k} dt + \dfrac1{2} \int_0^1 t^{4k+2} dt - \dfrac1{6} \int_0^1 t^{4k+3} dt \right)$$ $$\sum_{k=0}^{\infty} t^{4k+1} = \dfrac{t}{1-t^4}$$ $$\sum_{k=0}^{\infty} t^{4k} = \dfrac1{1-t^4}$$ $$\sum_{k=0}^{\infty} t^{4k+2} = \dfrac{t^2}{1-t^4}$$ $$\sum_{k=0}^{\infty} t^{4k+3} = \dfrac{t^3}{1-t^4}$$ $$S = \int_0^1 \left(-\dfrac12 \dfrac{t}{1-t^4} + \dfrac16 \dfrac1{1-t^4} + \dfrac12 \dfrac{t^2}{1-t^4} - \dfrac1{6} \dfrac{t^3}{1-t^4} \right) dt$$ $$g(t) = -\dfrac12 \dfrac{t}{1-t^4} + \dfrac16 \dfrac1{1-t^4} + \dfrac12 \dfrac{t^2}{1-t^4} - \dfrac1{6} \dfrac{t^3}{1-t^4}$$ $$6g(t) = \dfrac{1-3t+3t^2 - t^3}{(1-t^4)} = \dfrac{(1-t)^3}{1-t^4} = \dfrac{1-2t+t^2}{(1+t)(1+t^2)}$$ $$6g(t) = \dfrac2{t+1} - \dfrac1{1+t^2} - \dfrac{t}{1+t^2}$$ $$\int_0^1 6 g(t) dt = \int_0^1 \dfrac{2 dt}{1+t} - \int_0^1 \dfrac1{1+t^2} - \int_0^1 \dfrac{t}{1+t^2} = 2 \log(2) - \dfrac{\pi}4 - \dfrac12 \log(2) = \dfrac32 \log(2) - \dfrac{\pi}4$$ Hence, $$S = \int_0^1 g(t) dt = \dfrac14 \log(2) - \dfrac{\pi}{24}$$

EDIT

For what it is worth, in the first step if we write \begin{align} f(k) & = \dfrac1{(4k+1)(4k+2)(4k+3)(4k+4)}\\ & = - \dfrac14 \dfrac1{2k+1} + \dfrac16 \dfrac1{4k+1} + \dfrac12 \dfrac1{4k+3} - \dfrac1{24} \dfrac1{k+1}\\ & = -\dfrac14 \int_0^1 t^{2k} dt + \dfrac16 \int_0^1 t^{4k} dt + \dfrac1{2} \int_0^1 t^{4k+2} dt - \dfrac1{24} \int_0^1 t^k dt \end{align} not surprising you get an incorrect answer.

$$f(k) = -\dfrac14 \int_0^1 t^{2k} dt + \dfrac16 \int_0^1 t^{4k} dt + \dfrac1{2} \int_0^1 t^{4k+2} dt - \dfrac1{24} \int_0^1 t^k dt$$ $$S = \sum_{k=0}^{\infty} f(k) = \sum_{k=0}^{\infty} \left(-\dfrac14 \int_0^1 t^{2k} dt + \dfrac16 \int_0^1 t^{4k} dt + \dfrac1{2} \int_0^1 t^{4k+2} dt - \dfrac1{24} \int_0^1 t^k dt \right)$$ $$\sum_{k=0}^{\infty} t^{2k} = \dfrac1{1-t^2}$$ $$\sum_{k=0}^{\infty} t^{4k} = \dfrac1{1-t^4}$$ $$\sum_{k=0}^{\infty} t^{4k+2} = \dfrac{t^2}{1-t^2}$$ $$\sum_{k=0}^{\infty} t^{k} = \dfrac1{1-t}$$ $$S = \int_0^1 \left(-\dfrac14 \dfrac1{1-t^2} + \dfrac16 \dfrac1{1-t^4} + \dfrac12 \dfrac{t^2}{1-t^4} - \dfrac1{24} \dfrac1{1-t} \right) dt = \int_0^1 g(t) dt$$ $$g(t) = -\dfrac14 \dfrac1{1-t^2} + \dfrac16 \dfrac1{1-t^4} + \dfrac12 \dfrac{t^2}{1-t^4} - \dfrac1{24} \dfrac1{1-t}$$ $$g(t) = -\dfrac14 \dfrac1{1-t^2} + \dfrac16 \dfrac{3t^2+1}{1-t^4} - \dfrac1{24} \dfrac1{1-t} = -\dfrac1{24} \dfrac{t+7}{1-t^2} + \dfrac16 \dfrac{3t^2+1}{1-t^4}$$ $$g(t) = \dfrac1{24} \dfrac{12t^2+4 - (1+t^2)(7+t)}{1-t^4} = \dfrac1{24} \dfrac{-3-t+5t^2-t^3}{1-t^4} = \dfrac1{24} \dfrac{(1-t)(t^2-4t-3)}{1-t^4}$$ $$24 g(t) = \dfrac{t^2 - 4t-3}{(1+t)(1+t^2)} = \dfrac{t^2 +1 - 4t-4}{(1+t)(1+t^2)} = \dfrac{t^2+1}{(1+t)(1+t^2)} - 4 \dfrac{(1+t)}{(1+t)(1+t^2)}$$ $$24g(t) = \dfrac1{1+t} - \dfrac4{1+t^2}$$ Hence, $$S = \int_0^1 g(t) dt = \dfrac1{24} \left(\int_0^1 \dfrac{dt}{1+t} -4 \int_0^1 \dfrac{dt}{1+t^2}\right) = \dfrac1{24} \left( \left. \log(1+t) \right \vert_{t=0}^{t=1} - 4 \left. \arctan(t) \right \vert_{t=0}^{t=1}\right)$$ $$S = \dfrac1{24} \left( \log(2) - 4 \dfrac{\pi}4\right)$$

-

Start with the Mercator series $$\sum_{k=0}^\infty\frac{(-1)^k}{k+1}=\log(2)$$ and Gregory's series $$\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=\frac\pi4$$ The Heaviside Method yields \begin{align} &\frac6{(4k+1)(4k+2)(4k+3)(4k+4)}\\ &=\frac1{4k+1}-\frac3{4k+2}+\frac3{4k+3}-\frac1{4k+4}\\ &=\color{#C00000}{\left(\frac2{4k+1}-\frac2{4k+2}+\frac2{4k+3}-\frac2{4k+4}\right)}\\ &-\,\color{#00A000}{\left(\frac1{4k+1}-\frac1{4k+3}\right)}-\color{#0000FF}{\left(\frac1{4k+2}-\frac1{4k+4}\right)} \end{align} Note that the parts in red, green, and blue are all $O\left(\frac1{k^2}\right)$, so their sums converge absolutely.

Thus, \begin{align} \sum_{k=0}^\infty\frac6{(4k+1)(4k+2)(4k+3)(4k+4)} &=\color{#C00000}{2\log(2)}-\color{#00A000}{\frac\pi4}-\color{#0000FF}{\frac12\log(2)}\\ &=\frac32\log(2)-\frac\pi4 \end{align} Dividing by $6$ yields $$\sum_{k=0}^\infty\frac1{(4k+1)(4k+2)(4k+3)(4k+4)}=\frac14\log(2)-\frac\pi{24}$$

-