# Math Help - Ad hoc summation of series

1. ## Ad hoc summation of series

Hello,

is there a way of ad hoc summing 1/(1*2*3*4) + 1/(5*6*7*8) + ... without getting into the theory of Polygamma functions?

Thanks

2. You could look at the telescopic series

$\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n(n+1)(n+2)(n+3)} = \dfrac{1}{6}\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n+3}\right) - \dfrac{1}{2} \sum_{n=1}^{\infty} \left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)$

Edit: Sorry wrong series

3. Hello, Heirot!

is there a way of ad hoc summing: . $S \;=\;\dfrac{1}{1\cdot2\cdot3\cdot4} + \dfrac{1}{5\cdot6\cdot7\cdot8} + \hdots$

without getting into the theory of Polygamma functions?

The general term is: . $a_n \;=\;\dfrac{1}{n(n+1)(n+2)(n+3)}$

Partial fractions: . $\displaystyle \frac{1}{n(n+1)(n+2)(n+3)} \;=\;\frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} + \frac{D}{n+3}$

We find that: . $A = \frac{1}{6},\;B = -\frac{1}{2},\;C = \frac{1}{2},\;D = -\frac{1}{6}$

Hence: . $\displaystyle a_n \;=\;\frac{\frac{1}{6}}{n} - \frac{\frac{1}{2}}{n+1} + \frac{\frac{1}{2}}{n+2} - \frac{\frac{1}{6}}{n+3} \;=\;\frac{1}{6}\left(\frac{1}{n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)$

Then: . $\displaystyle S \;=\;\frac{1}{6}\sum^{\infty}_{n=1}\left(\frac{1}{ n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)$

Crank out several terms:

$\displaystyle S \;=\;\frac{1}{6}\bigg[\left(1 - \frac{3}{2} + \frac{3}{3} - \frac{1}{4}\right) + \left(\frac{1}{2} - \frac{3}{3} + \frac{3}{4} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6}\right)$

. . . . . . . $\displaystyle + \left(\frac{1}{4} - \frac{3}{5} + \frac{3}{6} - \frac{1}{7}\right) + \left(\frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8}\right) + \left(\frac{1}{6} - \frac{3}{7} + \frac{3}{8} - \frac{1}{9}\right)$

. . . . . . . . . $\displaystyle + \left(\frac{1}{7} - \frac{3}{8} + \frac{3}{9} - \frac{1}{10}\right) + \left(\frac{1}{8} - \frac{3}{9} + \frac{3}{10} - \frac{1}{11}\right) + \hdots \bigg]$

We find that nearly all the terms cancel out.

And we are left with: . $\displaystyle S \;=\;\frac{1}{6}\left(1 - \frac{3}{2} + \frac{1}{2} + \frac{1}{3}\right) \;=\;\frac{1}{18}$

4. The general term is in fact a_n = 1 / (4n (4n-1) (4n-2) (4n-3)) = (4(n-1))! /(4n)! and this it what complicates the sum. But thanks for trying, guys!

5. Originally Posted by Heirot
... But thanks for trying, guys!
What do you mean by that? They both are correct.

6. Originally Posted by chiph588@
What do you mean by that? They both are correct.
$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+ 2)(n+3)}=\frac{1}{1(2)3(4)}+\frac{1}{2(3)4(5)}+\fr ac{1}{3(4)5(6)}+....$

$\displaystyle \dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n$
then replace $x$ with $x^4$ then

$\displaystyle \dfrac{1}{1-x^4} = \sum_{n=0}^{\infty} x^{4n}$

integrate 4 times noting that the constants of integration are found by setting $x = 0$ into both sides of the equation. Then taking the limit as $x \to 1$ gives your answer

Spoiler:
$\text{I got }\;S = \dfrac{1}{4} \ln 2 - \dfrac{\pi}{24}$

8. \begin{aligned} \displaystyle \sum_{n=1}^\infty \frac1{4n(4n+1)(4n+2)(4n+3)} &= \sum_{n=1}^\infty\left(\frac{1/6}{4n}-\frac{1/2}{4n+1}+\frac{1/2}{4n+2}-\frac{1/6}{4n+3}\right)\\
&= \int_0^1\sum_{n=1}^\infty \left(\frac16x^{4n-1} -\frac12x^{4n}+\frac12x^{4n+1}-\frac16x^{4n-2}\right) dx\\
&= \int_0^1\left(-\frac16\frac{x^3}{x^4-1}+\frac12\frac{x^4}{x^4-1}-\frac12\frac{x^5}{x^4-1}+\frac16\frac{x^6}{x^4-1}\right)dx\\
&= -\frac16\int_0^1\left(\frac{x^3}{x^4-1}-\frac1{4(x-1)}\right)dx+\frac12\int_0^1\left(\frac{x^4}{x^4-1}-\frac1{4(x-1)}\right)dx\\&-\frac12\int_0^1\left(\frac{x^5}{x^4-1}-\frac1{4(x-1)}\right)dx+\frac16\int_0^1\left(\frac{x^6}{x^4-1}-\frac1{4(x-1)}\right)dx\\
&=-\frac1{12}\log2 +\frac12\left(1-\frac\pi8-\frac{\log2}4\right)-\frac14+\frac16\left(\frac1{24}(8+3\pi-6\log2)\right)\\
&= \frac1{72}(22-3\pi-18\log2)
\end{aligned}

Note, those four integrals are a pain though...

Edit: I did the wrong series! Your answer can be derived very similarly though..

9. You might need another series or a shift.

10. $\frac{1}{4n(4n-1)(4n-2)(4n-3)}=\frac{1}{8}(\frac{4}{3(4n-3)}+\frac{4}{4n-1}-\frac{2}{2n-1}-\frac{1}{3n})$

11. We may write $S = \displaystyle\sum_{n\geq 0}{\frac{1}{(4\cdot n + 1)...(4\cdot n + 4)}} = \frac{1}{6}\cdot \sum_{n\geq 0}{\text{B}(4n+1,4)}$ where $\text{B}(x,y)$ is the beta function.

Thus $6\cdot S = \displaystyle\sum_{n\geq 0}{\int_0^1{x^{4n}\cdot (1-x)^3}dx} = \int_0^1\left(\sum_{n\geq 0}x^{4n}\right)\cdot (1-x)^3dx = \int_0^1{\frac{(1-x)^3}{1-x^4}dx}$

Expanding the numerator: $6\cdot S = \displaystyle\int_0^1{\frac{(1-x)^2}{(1+x)\cdot (1+x^2)}dx}= \log(2) - \int_0^1{\frac{2x}{(1+x^2)}\cdot \frac{dx}{1+x}}$

Note that : $\displaystyle\frac{2x}{(1+x)\cdot (1+x^2)} = \displaystyle\frac{1+x}{1+x^2} - \frac{1}{1+x}$

And integrate.

12. *ignore this post... I misread the summation the OP posted*

13. An alternative. Let:

$S=\displaystyle\sum_{n=1}^{+\infty}a_n$

the sum of the given series. Using Soroban's decomposition:

$a_n=b_n+c_n,\quad b_n=\dfrac{1}{6}\left(\dfrac{1}{n}-\dfrac{1}{n+3}\right),\quad c_n=\dfrac{1}{2}\left(\dfrac{1}{n+2}-\dfrac{1}{n+1}\right)$

Now we use the well known property:

$\displaystyle\sum_{n=1}^{+\infty}(\varphi(n+q)-\varphi(n))=q\displaystyle\lim_{n \to{+}\infty}{\varphi (n)}-[ \varphi(1)+ \varphi (2)+\ldots + \varphi (q)]$

we inmediately obtain:

$S_1=\displaystyle\sum_{n=1}^{+\infty}b_n=\dfrac{11 }{36},\quad S_2=\displaystyle\sum_{n=1}^{+\infty}c_n=-\dfrac{1}{4}$

The sum of the given series is:

$S=S_1+S_2=\dfrac{11}{36}-\dfrac{1}{4}=\dfrac{1}{18}$

Fernando Revilla

Edited: What a strange thread! . I used Soroban's series instead of OP's series.

14. Why a_n=B(4n+1,4) ?

15. Originally Posted by Also sprach Zarathustra
Why a_n=B(4n+1,4) ?
It's because $\text{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$.

See here.

Page 1 of 2 12 Last