1. ## series

Challenge Problem:

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

Moderator editor: Approved Challenge question.

2. I am not sure my calculations are correct , please check it

We have

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

Therefore ,

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

your series should be equal to

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

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

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

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

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

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

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

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

The series

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

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

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

3. Originally Posted by Random Variable
Challenge Problem:

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

Moderator editor: Approved Challenge question.
This is $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)}$. Let us solve a more interesting sum.

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

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

So, remembering our definitions this may be written as

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

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

And remembering the formula for a geometeric sum we have that

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

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

Thus, $S_3=\frac{1}{18}$

4. Originally Posted by Drexel28
This is $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)}$.
No, it's not!

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

your series should be equal to

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

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

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

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

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

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

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

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

The series

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

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

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

6. Originally Posted by Bruno J.
No, it's not!
Haha, wow that was a little mistake. I can remedy it anyways...

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

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

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

So,

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

Switching integral and summation we get

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

Where the star indicates integration by method of partial fractions.

So, $S=\frac{1}{24}\left(6\ln(2)-\pi\right)$

7. Just for generalizations sake...

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

Then, using the exact same method

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

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

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

Wonder if anything meaningful lies in that beast

8. I expressed each term as a quadruple integral, reversed the order of integration, and then summmed up the terms.

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

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

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

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

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

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

$= \frac{1}{6} \Big( 2 \ln 2 - \frac{\pi}{4} - \frac{\ln 2}{2} \Big) = \frac{\ln 2}{4} - \frac{\pi}{24}$

9. Hi

I used another method, maybe less tricky

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

Let

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

The sum is equal to $f(1)$

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

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

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

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

(using the fact that $(1-x)^3 \ln(1-x)$ has 0 as limit when x approaches 1)

10. How did you go from the first line to the second line? My brain isn't working.

11. 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 $3 = 1 + 2$