# Thread: [SOLVED] Find sum of infinite series?

1. ## [SOLVED] Find sum of infinite series?

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)

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

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

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

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}}

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

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

We have
\displaystyle \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}

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

So
\displaystyle \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}

\displaystyle \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 :
\displaystyle \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}
$\displaystyle =\frac{1}{\ln(2)} \left( \sum_{n=1}^\infty \frac{1}{(n-1)!}+\frac 12 \sum_{n=0}^\infty \frac{(\frac 12)^n}{n!}\right)$
$\displaystyle =\frac{1}{\ln(2)} \left( \sum_{n=0}^\infty \frac{1}{n!}+\frac 12 \sum_{n=0}^\infty \frac{(\frac 12)^n}{n!}\right)$

Now, recall this series : $\displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}=e^x$
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 :/

6. Thanks Moo!

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

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

7. Originally Posted by fardeen_gen
Thanks Moo!

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

I don't see the interest either. But we got to do what we got to do
Hmm I think it's rather $\displaystyle \frac{2e+\sqrt{{\color{red}e}}}{2 \ln(2)}$, isn't it ?

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