Abstract Nonsense

Crushing one theorem at a time

Stimulating Sum


Problem: Compute \displaystyle S=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{5\cdot6\cdot7\cdot8}+\cdots

Solution: Note that

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

But, a little manipulation shows this is equal to

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

Thus,

\displaystyle 4! S=\sum_{n=0}^{\infty}\frac{(4n+5)4!(4n)!}{(4n+5)!}=\sum_{n=0}^{\infty}\frac{(4n+5)\Gamma(4n+1)\Gamma(4+1)}{\Gamma(4n+6)}

Which upon remembering some of the definitions is equal to

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

Where B(x,y) is the Beta function defined by

\displaystyle B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}=\int_0^1 t^{x-1}(1-t)^{y-1}dt

So,

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

Now, ignoring issues of uniform convergence (it is justifiable)

\displaystyle 4!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

But, with a little bit of elbow grease in the partial fractions department this is equal to

4!S=6\ln(2)-\pi

Thus,

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

More generally, it can be shown using exactly the same method that if

\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

That

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

Which when \alpha=1 has the very nice solution

\displaystyle S=\frac{1}{\beta\beta!}

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May 21, 2010 - Posted by | Analysis, Computations, Fun Problems, Munkres

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