Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I wanted to know how can I start to find the sum of the series:

$$\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}=\frac{1}{4!}+\frac{4!}{8!}+\frac{8!}{12!}\cdots$$

I am having no clue.

Thanks.

share|improve this question
    
Can you please cross check the numerator of the first term? –  lab bhattacharjee Jul 26 '13 at 12:47
1  
@lab, numerator is zero-factorial, seems right to me. –  Gerry Myerson Jul 26 '13 at 12:59
    
@GerryMyerson, I thought the problem to be of type simple exponential type, so could not match the pattern:) –  lab bhattacharjee Jul 26 '13 at 13:02
add comment

4 Answers

up vote 10 down vote accepted

Not nearly as impressive as the other answer, but elementary:

$$\begin{align} \frac{(4n)!}{(4n+4)!} &= \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)}\\ &= \left(\frac{1}{4n+1} - \frac{1}{4n+2}\right)\left(\frac{1}{4n+3} - \frac{1}{4n+4}\right)\\ &= \frac{1}{6} \frac{1}{4n+1} - \frac{1}{2}\frac{1}{4n+2} + \frac{1}{2}\frac{1}{4n+3} - \frac{1}{6}\frac{1}{4n+4}\\ &= \frac{1}{3}\left(\frac{1}{4n+1} - \frac{1}{4n+2} + \frac{1}{4n+3} - \frac{1}{4n+4}\right) - \frac{1}{6}\left(\frac{1}{4n+2} - \frac{1}{4n+4}\right) - \frac{1}{6} \left(\frac{1}{4n+1} - \frac{1}{4n+3}\right) \end{align}$$

It is well-known that

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

and the first parenthesis comprises four consecutive terms of that series, with no overlap, so from that we obtain $\frac{\log 2}{3}$. From the second parentheses, we can pull out a factor of $\frac12$ from both terms, then we obtain

$$\frac{1}{12} \left(\frac{1}{2n+1} - \frac{1}{2n+2}\right)$$

which comprises two consecutive terms of the $\log 2$ series, again without overlap, so together these two yield

$$\left(\frac{1}{3} - \frac{1}{12}\right)\log 2 = \frac{\log 2}{4}.$$

Another well-known series is Leibniz series

$$\frac{\pi}{4} = \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}$$

and the last parenthesis comprises two consecutive terms of that, once again without overlap.

Since all parenthesised terms are dominated by $\frac{1}{n^2}$, we can split and rearrange to obtain

$$\begin{align} \sum_{n = 0}^\infty \frac{(4n)!}{(4n+4)!} &= \frac13 \sum_{n = 0}^\infty \left(\frac{1}{4n+1} - \frac{1}{4n+2} + \frac{1}{4n+3} - \frac{1}{4n+4}\right)\\ &\quad -\frac{1}{12}\sum_{n=0}^\infty \left(\frac{1}{2n+1} - \frac{1}{2n+2}\right)\\ &\quad - \frac16 \sum_{n=0}^\infty \left(\frac{1}{4n+1} - \frac{1}{4n+3}\right)\\ &= \frac{\log 2}{3} - \frac{\log 2}{12} - \frac16\frac{\pi}{4} = \frac{\log 2}{4} - \frac{\pi}{24}. \end{align}$$

share|improve this answer
    
Mine was too similar. I've deleted it. –  robjohn Jul 26 '13 at 17:34
add comment

$$ \begin{aligned} \sum_{n\geq 0}\frac{(4n)!}{(4n+4)!} & =\sum_{n\geq 0} \frac{\Gamma(4n+1)}{\Gamma(4n+5)} \\& =\sum_{n\geq 0}\frac{1}{6}\mathrm{B}(4n+1, \,4) \\& = \frac{1}{6}\sum_{n\geq 0}\int_{0}^{1}x^{4n}(1-x)^3\; dx\\& = \frac{1}{6}\int_{0}^{1} \frac{(1-x)^2}{(1+x^2)(1+x)} \; dx \\& = \frac{\ln 2}{4}-\frac{\pi}{24} \end{aligned}$$

share|improve this answer
1  
What does $\Gamma$ signify? :) –  mikhailcazi Jul 26 '13 at 13:01
1  
@mikhailcazi Gamma function. In the second line is the related Beta function –  L. F. Jul 26 '13 at 13:03
7  
(+1) Nice answer. –  Mhenni Benghorbal Jul 26 '13 at 13:03
    
@MhenniBenghorbal Thank you! –  L. F. Jul 26 '13 at 13:24
2  
Forgive me for asking, but what college course would I take to learn how to answer this question in the same way? –  Caleb Jares Jul 26 '13 at 17:49
add comment

$$\begin{align*} \sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!} &=\sum_{n=0}^\infty\frac{1}{(4n+1)\times(4n+2)\times(4n+3)\times(4n+4)} \\& \end{align*}$$ And then decompose in simple element !

Use a fixed $N$ and evaluate the sum going from $0$ to $N$ by putting the 4 sums to the same start value and the same end value. Then everything will disapear.

share|improve this answer
add comment

$$\begin{align*} \sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!} &=\sum_{n=0}^\infty\frac{1}{(4n+1)\times(4n+2)\times(4n+3)\times(4n+4)} \\&=\frac{1}{8}\sum_{n=0}^\infty\frac{1}{32 n^4+80 n^3+70 n^2+25 n+3} \end{align*}$$

Now that we have replaced the nasty factorials with a nice, clean, rational function, we can use a more general procedure from here.

Also, if all you need to do is prove convergence, the comparison test is the simplest conclusive test.

If you want some additional stuff, here are the Wolfram Alpha queries for the equations: $\sum_{n=0}^\infty\frac{(4n)!}{(4n+4)!}$, $\frac{1}{8}\sum_{n=0}^\infty\frac{1}{32 n^4+80 n^3+70 n^2+25 n+3}$

share|improve this answer
    
Why am I downvoted? –  AJMansfield Nov 24 '13 at 21:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.