Ad hoc summation of series

**Heirot** · January 2nd 2011, 06:52 AM

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

**Danny** · January 2nd 2011, 07:12 AM

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

**Soroban** · January 2nd 2011, 07:38 AM

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

**Heirot** · January 2nd 2011, 07:47 AM

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!

**chiph588@** · January 2nd 2011, 09:22 AM

Originally Posted by Heirot

... But thanks for trying, guys!

What do you mean by that? They both are correct.

**Archie Meade** · January 2nd 2011, 10:02 AM

Originally Posted by chiph588@

What do you mean by that? They both are correct.

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+ 2)(n+3)}=\frac{1}{1(2)3(4)}+\frac{1}{2(3)4(5)}+\fr ac{1}{3(4)5(6)}+....$

**Danny** · January 2nd 2011, 10:55 AM

IF you start with the geometric power series

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

**chiph588@** · January 2nd 2011, 11:03 AM

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

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

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

**Danny** · January 2nd 2011, 11:12 AM

You might need another series or a shift.

**Also sprach Zarathustra** · January 2nd 2011, 11:15 AM

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

**PaulRS** · January 2nd 2011, 11:16 AM

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.

**Unbeatable0** · January 2nd 2011, 11:44 AM

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

**FernandoRevilla** · January 2nd 2011, 12:00 PM

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.

**Also sprach Zarathustra** · January 2nd 2011, 12:05 PM

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

**chiph588@** · January 2nd 2011, 12:16 PM

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.

Thread: Ad hoc summation of series

LinkBack

Thread Tools

Display

Ad hoc summation of series

Similar Math Help Forum Discussions

Summation of a series

Summation of a Series

summation of a series

Summation of the series

Need help with summation/series.

Search Tags