# Thread: Sum of the series.

1. ## Sum of the series.

Hi! I need help in finding the sum of the series:

I would appreciate any kind of help here, since I don't even know how to start.

2. Originally Posted by Rist
Hi! I need help in finding the sum of the series:

I would appreciate any kind of help here, since I don't even know how to start.

-Dan

3. Originally Posted by Rist
Hi! I need help in finding the sum of the series:

I would appreciate any kind of help here, since I don't even know how to start.
For the first one, both x=0 and x=1 give divergent series

as for the second one I am not sure that can be easily summed

4. The second is easily summed: $x = 12\quad \Rightarrow \quad \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)2^n }}}$

5. I think that x=0,1 means x=1/10 which gives a convergent series. Anyway I'll solve it for general.

$S = \sum_{n=0}^{\infty} \frac{x^n}{n+1}$

$x\cdot S = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}$

$x\cdot S = \int \sum_{n=0}^{\infty} \frac{\partial}{\partial x} \frac{x^{n+1}}{n+1}~dx$

$x\cdot S = \int \sum_{n=0}^{\infty} x^n~dx$

$x\cdot S = \int \frac{1}{1-x} ~dx$

$x\cdot S = - \ln (1-x) + C$

We can find C using,

$x\cdot S = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = - \ln (1-x) + C$

Use $x = 0$, then
$0=-\ln (1-0)+C$

$C=0$

So finally,

$x\cdot S = - \ln (1-x)$

$S = - \frac{\ln (1-x)}{x}$

$\sum_{n=0}^{\infty} \frac{x^n}{n+1} = - \frac{\ln (1-x)}{x}$

---

Assuming $x=0.1$ (this is the x in the original question, so 1-6x makes 0.4),

$S = - \frac{\ln (1-0.4)}{0.4} \approx 1.27706$

6. $\sum\limits_{n\,=\,1}^{\infty }{\frac{1}{2^{n}n(n+1)}}=\sum\limits_{n\,=\,1}^{\i nfty }{\frac{(n-1)!}{2^{n}(n+1)!}}=\sum\limits_{n\,=\,1}^{\infty }{\frac{\Gamma (n)\Gamma (2)}{2^{n}\Gamma (n+2)}}.$

Hence the sum equals,

$\sum\limits_{n\,=\,1}^{\infty }{\int_{0}^{1}{\frac{x^{n-1}(1-x)}{2^{n}}\,dx}}=\int_{0}^{1}{\left\{ \frac{1-x}{x}\sum\limits_{n\,=\,1}^{\infty }{\left( \frac{x}{2} \right)^{n}} \right\}\,dx}=\int_{0}^{1}{\frac{1-x}{2-x}\,dx}$

and the conclusion follows. (This sum can be tackled from many ways.)

7. Originally Posted by Plato
The second is easily summed: $x = 12\quad \Rightarrow \quad \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)2^n }}}$
Originally Posted by Krizalid
$\sum\limits_{n\,=\,1}^{\infty }{\frac{1}{2^{n}n(n+1)}}=\sum\limits_{n\,=\,1}^{\i nfty }{\frac{(n-1)!}{2^{n}(n+1)!}}=\sum\limits_{n\,=\,1}^{\infty }{\frac{\Gamma (n)\Gamma (2)}{2^{n}\Gamma (n+2)}}.$

Hence the sum equals,

$\sum\limits_{n\,=\,1}^{\infty }{\int_{0}^{1}{\frac{x^{n-1}(1-x)}{2^{n}}\,dx}}=\int_{0}^{1}{\left\{ \frac{1-x}{x}\sum\limits_{n\,=\,1}^{\infty }{\left( \frac{x}{2} \right)^{n}} \right\}\,dx}=\int_{0}^{1}{\frac{1-x}{2-x}\,dx}$

and the conclusion follows. (This sum can be tackled from many ways.)
That's what you call easily summed?

8. Originally Posted by Mathstud28
That's what you call easily summed?
For Krizalid, that's pretty straightforward. He's not the "Inte-killer" for nothing!

-Dan

9. Originally Posted by Mathstud28

That's what you call easily summed?
I don't say that this is called "easily summed." Why do you say that?

10. Originally Posted by Krizalid
I don't say that this is called "easily summed." Why do you say that?
Plato: The second is easily summed:

^^

11. Originally Posted by Mathstud28
That's what you call easily summed?
Since $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$:

$S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)2^n} = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^n}$

$= \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - 2 \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^{n+1}}$ ........

12. Originally Posted by mr fantastic
Since $\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$:

$S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)2^n} = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^n} = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - 2 \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^{n+1}}$ ........
Wow haha...this has really turned into a big deal. I was stating that for most people this wouldn't be the easiest to sum series.

13. Ahhh well, but, why are you talkin' to me for, if Plato said so.

14. Originally Posted by Krizalid
Ahhh well, but, why are you talkin' to me for, if Plato said so.
I quoted you for your solution so that when I said "That whats you call easy to sum" to Plato it had dramatic effect

Note to self: Don't say things are easy

15. *Ahem*
Originally Posted by Mathstud28
[snip]
as for the second one I am not sure that can be easily summed
Plato remarked that it was.

I doubt Krizalid's approach was what Plato had in mind (which in no way is meant to diminish K's solution).

My post is what Plato might have had in mind (I won't presume that it was). I think it's evident that Plato's remark was both reasonable and accurate.

Originally Posted by Mathstud28
Wow haha...this has really turned into a big deal. I was stating that for most people this wouldn't be the easiest to sum series.
True, since most people don't study to the level that this question is contexted within.

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