series

May 2009
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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.
 
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simplependulum

MHF Hall of Honor
Jan 2009
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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

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

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

\(\displaystyle = \frac{1}{24} ( 6\ln{2} - \pi ) \)
 
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Drexel28

MHF Hall of Honor
Nov 2009
4,563
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Berkeley, California
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

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

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, \(\displaystyle S_3=\frac{1}{18}\)
 

Bruno J.

MHF Hall of Honor
Jun 2009
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498
Canada
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!
 

Drexel28

MHF Hall of Honor
Nov 2009
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1,566
Berkeley, California
No, it's not! (Thinking)
Haha, wow that was a little mistake. I can remedy it anyways...

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

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

Drexel28

MHF Hall of Honor
Nov 2009
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1,566
Berkeley, California
Just for generalizations sake...

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

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
 
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May 2009
959
362
I expressed each term as a quadruple integral, reversed the order of integration, and then summmed up the terms.


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

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


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

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

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

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

\(\displaystyle = \frac{1}{6} \Big( 2 \ln 2 - \frac{\pi}{4} - \frac{\ln 2}{2} \Big) = \frac{\ln 2}{4} - \frac{\pi}{24}\)
 
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Nov 2008
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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)
 
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