Find the exact sum of the series 1/(1⋅2⋅3⋅4)+1/(5⋅6⋅7⋅8)+....

[lfdahl]

Find the exact sum of the series:

$$S = \frac{1}{1\cdot 2\cdot 3\cdot 4}+\frac{1}{5 \cdot 6 \cdot 7 \cdot 8}+...$$

 

[MountEvariste]

Spoiler

First observe that the sum is $\displaystyle \sum_{k \ge 0} \frac{(4k)!}{(4k+4)!}$ and more crucially recall that $\displaystyle \text{B}(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$.
Write $\displaystyle \frac{(4k)!}{(4k+4)!} = \frac{\Gamma(4k+1)}{\Gamma(4k+5)} = \frac{\Gamma(4k+1)\Gamma(4)}{3! ~\Gamma(4k+5)} = \frac{1}{6}\text{B}(4k+1, 4) = \frac{1}{6}\int_0^1t^{4k}(1-t)^{3}\,{dt}$ so

$\displaystyle \sum_{k \ge 0} \frac{(4k)!}{(4k+4)!} = \sum_{k \ge 0} \frac{1}{6}\int_0^1t^{4k}(1-t)^{3}\,{dt} = \frac{1}{6}\int_0^1\sum_{k \ge 0} t^{4k}(1-t)^{3}\,{dt} = \frac{1}{6}\int_0^1 \frac{(1-t)^3}{1-t^4}\,{dt}.$

The integral easily evaluates to $\displaystyle \frac{1}{4} \left(\log(64)-\pi\right) $ so the value sought is $\displaystyle \frac{1}{24} \left(\log(64)-\pi\right).$

 

[lfdahl]

Spoiler

First observe that the sum is $\displaystyle \sum_{k \ge 0} \frac{(4k)!}{(4k+4)!}$ and more crucially recall that $\displaystyle \text{B}(x,y)=\frac{\Gamma(x)\,\Gamma(y)}{\Gamma(x+y)}$.
Write $\displaystyle \frac{(4k)!}{(4k+4)!} = \frac{\Gamma(4k+1)}{\Gamma(4k+5)} = \frac{\Gamma(4k+1)\Gamma(4)}{3! ~\Gamma(4k+5)} = \frac{1}{6}\text{B}(4k+1, 4) = \frac{1}{6}\int_0^1t^{4k}(1-t)^{3}\,{dt}$ so

$\displaystyle \sum_{k \ge 0} \frac{(4k)!}{(4k+4)!} = \sum_{k \ge 0} \frac{1}{6}\int_0^1t^{4k}(1-t)^{3}\,{dt} = \frac{1}{6}\int_0^1\sum_{k \ge 0} t^{4k}(1-t)^{3}\,{dt} = \frac{1}{6}\int_0^1 \frac{(1-t)^3}{1-t^4}\,{dt}.$

The integral easily evaluates to $\displaystyle \frac{1}{4} \left(\log(64)-\pi\right) $ so the value sought is $\displaystyle \frac{1}{24} \left(\log(64)-\pi\right).$

Well done, June29! Thankyou for your participation.

Spoiler

Maybe you will share with the forum, your evaluation of the last integral in your solution?

 

[MountEvariste]

Well done, June29! Thankyou for your participation.

Spoiler

Maybe you will share with the forum, your evaluation of the last integral in your solution?

Thank you for the wonderful problems you post!

Spoiler

\(\displaystyle \displaystyle \frac{(1-x)^3}{1-x^4} = \frac{(1-x)^3}{(1-x)(1+x)(1+x^2)} = \frac{(1-x)^2}{(1+x)(1+x^2)}\)

Write $(1-x)^2 = 1-2x+x^2 = 2(1+x^2)-(1+x)^2$

It becomes $ \displaystyle \frac{2(1+x^2)-(1+x)^2}{(1+x)(1+x^2)} = \frac{2}{1+x}-\frac{1+x}{1+x^2}$

or $\displaystyle \frac{2}{1+x}-\frac{1}{1+x^2}-\frac{1}{2}\frac{2x}{1+x^2}$, which readily integrates to

$\displaystyle 2 \log(1+x)-\arctan(x)-\frac{1}{2}\log(1+x^2)\bigg|_0^{1} = \frac{1}{4}(\log(64)-\pi). $

I'd be interested in your solution if it isn't the same as mine.

 

[lfdahl]

Thank you for the wonderful problems you post!

Spoiler

\(\displaystyle \displaystyle \frac{(1-x)^3}{1-x^4} = \frac{(1-x)^3}{(1-x)(1+x)(1+x^2)} = \frac{(1-x)^2}{(1+x)(1+x^2)}\)

Write $(1-x)^2 = 1-2x+x^2 = 2(1+x^2)-(1+x)^2$

It becomes $ \displaystyle \frac{2(1+x^2)-(1+x)^2}{(1+x)(1+x^2)} = \frac{2}{1+x}-\frac{1+x}{1+x^2}$

or $\displaystyle \frac{2}{1+x}-\frac{1}{1+x^2}-\frac{1}{2}\frac{2x}{1+x^2}$, which readily integrates to

$\displaystyle 2 \log(1+x)-\arctan(x)-\frac{1}{2}\log(1+x^2)\bigg|_0^{1} = \frac{1}{4}(\log(64)-\pi). $

I'd be interested in your solution if it isn't the same as mine.

Thankyou for your kind words, and for showing the last part of your solution! I guess, there is no need to post the suggested solution, since it is very close to yours!(Nod)

 

[MountEvariste]

Wait... I actually came across this problem years ago (IMC, right?)! (Giggle)

Spoiler

Generalisation: \(\displaystyle \displaystyle \sum_{k \ge 0 }\prod_{1 \le r \le j}\frac{1}{(n k+r)} = \frac{\sqrt{\pi}}{n(j-1)!}\sum_{0 \le r < j}\binom{j-1}{r}\frac{(-1)^r \Gamma( \frac{r+1}{n})}{\Gamma(\frac{r+1 }{n}+\frac{1}{2})} \)

Cause LHS can be written as $\displaystyle \frac{1}{(j-1)!}\int_{0}^{1}\frac{(1-t)^{j-1}}{1-t^n}\;{dt}$ (same as before).

We can also prove the OP via partial fractions alone, but it's quite laborious!