# series

• May 16th 2010, 07:48 PM
Random Variable
series
Challenge Problem:

Find the sum of $\displaystyle \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7 \cdot 8} + \frac{1}{9 \cdot 10 \cdot 11 \cdot 12} + ...$

Moderator editor: Approved Challenge question.
• May 16th 2010, 08:36 PM
simplependulum
I am not sure my calculations are correct , please check it

We have

$\displaystyle \prod_{k=0}^{n} ( x + k)^{-1} = \frac{1}{n!} \sum_{k=0}^n \frac{(-1)^k \binom{n}{k}}{ x + k}$

Therefore ,

$\displaystyle \frac{1}{ x(x+1)(x+2)(x+3) } = \frac{1}{6} \left( \frac{1}{x} - \frac{3}{x+1} + \frac{3}{x+2} - \frac{1}{x+3} \right )$

your series should be equal to

$\displaystyle \frac{1}{6} ( \frac{1}{1} - \frac{3}{2} + \frac{3}{3} - \frac{1}{4} +$ $\displaystyle \frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8} +$ $\displaystyle \frac{1}{9} - \frac{3}{10} + \frac{3}{11} - \frac{1}{12} + ...$

$\displaystyle = \frac{1}{6} [ ( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... ) - 2 ( \frac{1}{2} - \frac{1}{3} + \frac{1}{6} - \frac{1}{7} + .... )]$

$\displaystyle = \frac{1}{6} [ \ln{2} - 2 ( \int_0^1 ( x - x^2 + x^5 - x^6 + ... )~dx]$

$\displaystyle = \frac{1}{6} \left[ \ln{2} - 2 \left( \int_0^1 [x( 1 + x^4 + ...) - x^2 ( 1 + x^4+...)]~dx \right ) \right ]$

$\displaystyle = \frac{1}{6} [ \ln{2} - 2 \int_0^1 \frac{x(1-x)}{1 - x^4}~dx ]$

The integral
$\displaystyle \int_0^1 \frac{x(1-x)}{1 - x^4}~dx$

$\displaystyle = \frac{1}{2} \int_0^1 \left( \frac{x+1}{x^2+1} - \frac{1}{x+1} \right)~dx$

$\displaystyle = \frac{1}{2} ( \frac{\pi}{4} - \frac{\ln{2}}{2} )$

The series

$\displaystyle = \frac{1}{6} [ \ln{2} - \frac{\pi}{4} + \frac{\ln{2}}{2} ]$

$\displaystyle = \frac{1}{6} [ \frac{3}{2} \ln{2} - \frac{\pi}{4} ]$

$\displaystyle = \frac{1}{24} ( 6\ln{2} - \pi )$
• May 16th 2010, 08:40 PM
Drexel28
Quote:

Originally Posted by Random Variable
Challenge Problem:

Find the sum of $\displaystyle \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7 \cdot 8} + \frac{1}{9 \cdot 10 \cdot 11 \cdot 12} + ...$

Moderator editor: Approved Challenge question.

This is $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)}$. Let us solve a more interesting sum.

Let $\displaystyle S_\ell=\sum_{n=1}^{\infty}\frac{1}{f_\ell(n)}$ where $\displaystyle f_\ell(n)=\prod_{j=0}^{\ell}(n+j),\text{ }\ell\geqslant 1$.

Notice then that $\displaystyle S_\ell=\sum_{n=1}^{\infty}\frac{1}{f_\ell(n)}=\sum _{n=0}^{\infty}\frac{n!\ell!}{\ell!(n+\ell+1)!}$.

So, remembering our definitions this may be written as

$\displaystyle S_\ell=\sum_{n=0}^{\infty}\frac{\Gamma(n+1)\Gamma( \ell+1)}{\ell!\Gamma(n+\ell+2)}=\frac{1}{\ell!}\su m_{n=0}^{\infty}B(n+1,\ell+1)$.

Thus, $\displaystyle \ell! S_\ell=\sum_{n=0}^{\infty}\int_0^1 t^n(1-t)^\ell dt=\int_0^1 (1-t)^\ell\sum_{n=0}^{\infty}t^n dt$

And remembering the formula for a geometeric sum we have that

$\displaystyle \ell! S_\ell=\int_0^1(1-t)^{\ell-1}=\frac{1}{\ell}$.

Thus, $\displaystyle S_\ell=\frac{1}{\ell \ell!}$.

Thus, $\displaystyle S_3=\frac{1}{18}$
• May 16th 2010, 08:51 PM
Bruno J.
Quote:

Originally Posted by Drexel28
This is $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)}$.

No, it's not! (Thinking)
• May 16th 2010, 08:54 PM
Bruno J.
Quote:

Originally Posted by simplependulum
I am not sure my calculations are correct , please check it

your series should be equal to

$\displaystyle \frac{1}{6} ( \frac{1}{1} - \frac{3}{2} + \frac{3}{3} - \frac{1}{4} +$ $\displaystyle \frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8} +$ $\displaystyle \frac{1}{9} - \frac{3}{10} + \frac{3}{11} - \frac{1}{12} + ...$

$\displaystyle = \frac{1}{6} [ ( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... ) - 2 ( \frac{1}{2} - \frac{1}{3} + \frac{1}{6} - \frac{1}{7} + .... )]$

$\displaystyle = \frac{1}{6} [ \ln{2} - 2 ( \int_0^1 ( x - x^2 + x^5 - x^6 + ... )~dx]$

$\displaystyle = \frac{1}{6} \left[ \ln{2} - 2 \left( \int_0^1 [x( 1 + x^4 + ...) - x^2 ( 1 + x^4+...)]~dx \right ) \right ]$

$\displaystyle = \frac{1}{6} [ \ln{2} - 2 \int_0^1 \frac{x(1-x)}{1 - x^4}~dx ]$

The integral
$\displaystyle \int_0^1 \frac{x(1-x)}{1 - x^4}~dx$

$\displaystyle = \frac{1}{2} \int_0^1 \left( \frac{x+1}{x^2+1} - \frac{1}{x+1} \right)~dx$

$\displaystyle = \frac{1}{2} ( \frac{\pi}{4} - \frac{\ln{2}}{2} )$

The series

$\displaystyle = \frac{1}{6} [ \ln{2} - \frac{\pi}{4} + \frac{\ln{2}}{2} ]$

$\displaystyle = \frac{1}{6} [ \frac{3}{2} \ln{2} - \frac{\pi}{4} ]$

$\displaystyle = \frac{1}{24} ( 6\ln{2} - \pi )$

Mathematica agrees! Good job! Maybe the first step could be written more clearly (it's not obvious you used partial fractions!) but you did an impressive job there!
• May 16th 2010, 09:24 PM
Drexel28
Quote:

Originally Posted by Bruno J.
No, it's not! (Thinking)

Haha, wow that was a little mistake. I can remedy it anyways...

So, let $\displaystyle S=\sum_{n=0}^{\infty}\frac{1}{(4n+1)(4n+2)(4n+3)(4 n+4)}$

Using the exact same trick as my early proof we can transform this into

$\displaystyle S=\sum_{n=0}^{\infty}\frac{(4n+5)4!(4n!)}{4!(4n+5) !}=\frac{1}{4!}\sum_{n=0}^{\infty}(4n+5)B(4n+1,4+1 )$

So,

$\displaystyle 4!\cdot S=\sum_{n=0}^{\infty}(4n+5)\int_0^1 t^{4n}(1-t)^4 dt$

Switching integral and summation we get

$\displaystyle 4!\cdot S=\int_0^1 (1-t)^4\sum_{n=0}^{\infty}(4n+5)t^{4n}=\int_0^1\frac{ (1-t)^4(5-t^4)}{(1-t^4)^2}dt\overset{\color{red}(*)}{=}6\ln(2)-\pi$

Where the star indicates integration by method of partial fractions.

So, $\displaystyle S=\frac{1}{24}\left(6\ln(2)-\pi\right)$
• May 16th 2010, 09:41 PM
Drexel28
Just for generalizations sake...

Let $\displaystyle S(\alpha,\beta)=\sum_{n=0}^{\infty}\frac{1}{(\alph a n+1)\cdots(\alpha n+\beta)},\text{ }\alpha,\beta\in\mathbb{N},\text{ }\beta\geqslant 1$

Then, using the exact same method

$\displaystyle S(\alpha,\beta)=\frac{1}{\beta!}\sum_{n=0}^{\infty }(\alpha n+\beta+1)B(\alpha n +1,\beta +1)$

Or, $\displaystyle \beta!S(\alpha,\beta)=\int_0^1(1-t)^\beta\sum_{n=0}^{\infty}(\alpha n+\beta)t^{\alpha n}$

Or, $\displaystyle \beta !S(\alpha,\beta)=\int_0^1\frac{(1-t)^\beta\cdot(\alpha t^\alpha+\beta(1-t^\alpha))}{(1-t^\alpha)^2}dt$

Wonder if anything meaningful lies in that beast
• May 17th 2010, 01:30 AM
Random Variable
I expressed each term as a quadruple integral, reversed the order of integration, and then summmed up the terms.

$\displaystyle \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)} = \int_{0}^{1} \int^{z}_{0} \int^{y}_{0} \int^{x}_{0} w^{4n} \ dw \ dx \ dy \ dz$

$\displaystyle = \int_{0}^{1} \int^{1}_{w} \int^{1}_{x} \int^{1}_{y} w^{4n} \ dz \ dy \ dx \ dw = \int^{1}_{0} w^{4n} \Big(\frac{1}{6} - \frac{w}{2} + \frac{w^{2}}{2} - \frac{w^{3}}{6} \Big) \ dw$ = $\displaystyle = \frac{1}{6} \int^{1}_{0} w^{4n} (1-w)^{3} \ dw$

$\displaystyle \sum^{\infty}_{n=0} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)} = \frac{1}{6} \sum^{\infty}_{n=0} \int^{1}_{0} w^{4n}(1-w)^{3} \ dw$

$\displaystyle = \frac{1}{6}\int^{1}_{0} (1-w)^{3} \sum^{\infty}_{n=0} w^{4n}\ dw = \frac{1}{6} \int^{1}_{0} \frac{(1-w)^{3}}{1-w^{4}} \ dw$

$\displaystyle = \frac{1}{6} \int^{1}_{0} \frac{(1-w)^{3}}{(1-w)(1+w)(1+w^{2})} \ dw = \frac{1}{6} \int^{1}_{0} \frac{(1-w)^{2}}{(1+w)(1+w^{2})} \ dw$

$\displaystyle = \frac{1}{6} \int^{1}_{0} \Big(\frac{2}{1+w} - \frac{1}{1+w^{2}} - \frac{w}{1+w^{2}} \ \Big) \ dw = \frac{1}{6} \Big( 2 \ln (1+w) - \arctan w - \frac{\ln(1+w^{2})}{2} \Big) \Big|^{1}_{0}$

$\displaystyle = \frac{1}{6} \Big( 2 \ln 2 - \frac{\pi}{4} - \frac{\ln 2}{2} \Big) = \frac{\ln 2}{4} - \frac{\pi}{24}$
• May 17th 2010, 01:53 AM
running-gag
Hi

I used another method, maybe less tricky

$\displaystyle \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7 \cdot 8} + \frac{1}{9 \cdot 10 \cdot 11 \cdot 12} + ... = \sum_{k=0}^{+\infty} \frac{(4k)!}{(4k+4)!}$

Let

$\displaystyle f(x) = \sum_{k=0}^{+\infty} \frac{(4k)!}{(4k+4)!} x^{4k+4}$

The sum is equal to $\displaystyle f(1)$

The fourth derivative of f is $\displaystyle f^{(4)}(x) = \sum_{k=0}^{+\infty} x^{4k} = \frac{1}{1-x^4}$

The fourth anti-derivative of $\displaystyle f^{(4)}(x)$ is

$\displaystyle f(x) = \frac{(1+x)^3}{24} \ln(1+x) +\frac{(1-x)^3}{24} \ln(1-x) - \frac{3x^2-1}{24} \ln(1+x^2) + \frac{x^3-3x}{12} \arctan(x)$

And $\displaystyle f(1) = \frac{\ln(2)}{4} - \frac{\pi}{24}$

(using the fact that $\displaystyle (1-x)^3 \ln(1-x)$ has 0 as limit when x approaches 1)
• May 17th 2010, 01:59 AM
Random Variable
How did you go from the first line to the second line? My brain isn't working.
• May 17th 2010, 02:22 AM
simplependulum
Quote:

Originally Posted by Random Variable
How did you go from the first line to the second line? My brain isn't working.

I have put $\displaystyle 3 = 1 + 2$