[SOLVED] Find sum of infinite series?

**fardeen_gen** · April 23rd 2009, 05:34 AM

Find the sum of the following infinite series:

∑(n = 0 to ∞) (1/n![∑(k = 0 to n) ((k+1).∫(from 0 to 1) 2^(-(k +1)x) dx]8)

**fardeen_gen** · April 25th 2009, 09:55 PM

Is it something to do with binomial or A.P/G.P/H.P? I cannot make head or tail still.

**Moo** · April 26th 2009, 12:40 AM

Just a question : can you possibly learn latex ? ^^

**fardeen_gen** · April 27th 2009, 08:57 PM

$\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)]8}}$

Thanks for the push Moo!

Can anyone help now?

EDIT: Why is this not working(I want to make the square brackets big):

Code:

\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}

**Moo** · April 28th 2009, 01:58 AM

Originally Posted by fardeen_gen

$\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)]8}}$

Thanks for the push Moo!

Can anyone help now?

EDIT: Why is this not working(I want to make the square brackets big):
\sum\limits_{n = 0}^{ \infty } {\frac{1}{n!}\left[\sum\limits_{k = 0}^{n} {(k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8}}

$\sum\limits_{n = 0}^{ \infty } \frac{1}{n!}\left[\sum\limits_{k = 0}^{n} (k + 1)\left(\int_0^1 2^{-(k + 1)x}\ dx\right)\right]8$
I removed the red brackets

Anyway, for your problem... :

We have
$\begin{aligned} \int_0^1 2^{-(k+1)x} ~dx &=\int_0^1 \exp\left(-(k+1)x\ln(2)\right) ~dx \\ &=\left. -\frac{1}{(k+1)\ln(2)} \cdot \exp\left(-(k+1)x\ln(2)\right)\right|_0^1 \\ &=\left. -\frac{1}{(k+1)\ln(2)} \cdot 2^{-(k+1)x}\right|_0^1 \\ &=-\frac{1}{(k+1)\ln(2)} \cdot (2^{-(k+1)}-1) \end{aligned}$

$\int_0^1 2^{-(k+1)x} ~dx = \frac{1}{(k+1)\ln(2)} \cdot \left(1-\frac{1}{2^{k+1}}\right)$

So
$\begin{aligned} \sum_{k=0}^n (k+1)\int_0^1 2^{-(k+1)x} ~dx &=\frac{1}{\ln(2)} \sum_{k=0}^n 1-\frac{1}{2^{k+1}} \\ &=\frac{1}{\ln(2)} \left(\sum_{k=0}^n 1 -\sum_{k=0}^n \frac{1}{2^{k+1}}\right) \\ &=\frac{1}{\ln(2)} \left((n+1)-\frac 12 \cdot \frac{1-\frac{1}{2^{n+1}}}{1-\frac 12}\right) \\ &=\frac{1}{\ln(2)} \left((n+1)-\left(1-\frac{1}{2^{n+1}}\right)\right) \end{aligned}$

$\begin{aligned} \sum_{k=0}^n (k+1)\int_0^1 2^{-(k+1)x} ~dx &=\frac{1}{\ln(2)} \left(n+\frac{1}{2^{n+1}}\right) \end{aligned}$

Now, we have your initial series equal to :
$\begin{aligned} &=\frac{1}{\ln(2)} \sum_{n=0}^\infty \frac{\left(n+\frac{1}{2^{n+1}}\right)}{n!} \\ &=\frac{1}{\ln(2)} \left( \sum_{n=0}^\infty \frac{n}{n!}+\sum_{n=0}^\infty \frac{(\frac 12)^{n+1}}{n!}\right) \\ &=\frac{1}{\ln(2)} \left( \sum_{n=1}^\infty \frac{n}{n!}+\frac 12 \sum_{n=0}^\infty \frac{(\frac 12)^n}{n!}\right) \end{aligned}$
$=\frac{1}{\ln(2)} \left( \sum_{n=1}^\infty \frac{1}{(n-1)!}+\frac 12 \sum_{n=0}^\infty \frac{(\frac 12)^n}{n!}\right) $
$=\frac{1}{\ln(2)} \left( \sum_{n=0}^\infty \frac{1}{n!}+\frac 12 \sum_{n=0}^\infty \frac{(\frac 12)^n}{n!}\right)$

Now, recall this series : $\sum_{n=0}^\infty \frac{x^n}{n!}=e^x$
For the first one : x=1, for the second one, x=1/2

And it should be finished.

HOwever, I don't quite see the interest of calculating such a series :/

**fardeen_gen** · April 28th 2009, 04:50 AM

Thanks Moo!

The answer: $\frac{2e + \sqrt{2}}{2\ln2}$

HOwever, I don't quite see the interest of calculating such a series :/

I don't see the interest either. But we got to do what we got to do

EDIT: My typo

Answer: $\frac{2e + \sqrt{e}}{2\ln2}$

**Moo** · April 28th 2009, 09:33 AM

Originally Posted by fardeen_gen

Thanks Moo!

The answer: $\frac{2e + \sqrt{2}}{2\ln2}$

I don't see the interest either. But we got to do what we got to do

Hmm I think it's rather $\frac{2e+\sqrt{{\color{red}e}}}{2 \ln(2)}$ , isn't it ?

You're welcome. And you're right ^^

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