# Evaluate the summation

• Jul 8th 2006, 05:48 PM
Nichelle14
Evaluate the summation
Evaluate:

from 1 to infinity the summation of [n/2^n-1]

I tried to work it out for a small amounts of n to see if I can see anything happening. What should I do next?
• Jul 8th 2006, 10:32 PM
earboth
Quote:

Originally Posted by Nichelle14
Evaluate:

from 1 to infinity the summation of [n/2^n-1]

I tried to work it out for a small amounts of n to see if I can see anything happening. What should I do next?

Hello, Nichelle,

I'm puzzled, because it is not clear for me, what you mean. Do you mean:

$\frac{n}{2^n}-1\quad \mbox{or}\quad \frac{n}{2^n-1}\quad \mbox{or}\quad \frac{n}{2^{n-1}}$

Greetings

EB
• Jul 9th 2006, 03:33 AM
rgep
Here's a general technique that may help. I'll tackle the example $\frac{n}{2^{n-1}}$. The trick is to write down a function $f(x)$ for which the sum is a special value. Here we consider $f(x) = \sum_n \frac{n}{x^{n-1}}$, so that you want $f(2)$. We look at $x^{-2} f(x) = \sum_n \frac{n}{x^{n+1}}$ in order to be able to integrate term by term. So $x^{-2}f(x)$ is the derivative of $\sum_n \frac{-1}{x^n}$, which is $\frac{-1}{1-1/x} = \frac{x}{1-x} = 1+\frac{1}{1-x}$. Hence $x^{-2} f(x) = \frac{1}{(1-x)^2}$ and your sum is $f(2) = 4$.
• Jul 9th 2006, 04:59 AM
galactus
I'll try the $\sum_{n=1}^{\infty}\frac{n}{2^{n}}-1$

You can just use the closed form for a geometric series, namely, $\frac{1}{1-x}$

$\frac{1}{1-\frac{1}{2}}=2$

$\sum_{n=1}^{\infty}\frac{n}{2^{n-1}}$ can be treated similarly.

This is the same as:

$\frac{n}{2^{n}2^{-1}}$

$2\sum_{n=1}^{\infty}\frac{n}{2^{n}}=\frac{2}{1-\frac{1}{2}}=4$
• Jul 9th 2006, 05:26 AM
Soroban
Hello, Nichelle14!

I'll assume that the exponent is $n-1$.

Quote:

Evaluate: . $\sum^{\infty}_{n=1}\frac{n}{2^{n-1}}$

Here's an elementary approach . . .

We are given: . $S\;=\;1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \frac{5}{2^4} + \hdots$

Divide by 2: . $\frac{1}{2}S \;= \;\;\;\,\quad\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \frac{4}{2^4} + \hdots$

Subtract:. . . . $\frac{1}{2}S\;= \;1 + \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4} + \hdots$

The right side is a geometric series with $a = 1,\;r = \frac{1}{2}$
. . Hence, its sum is: . $\frac{1}{1 - \frac{1}{2}} \,= \,2$

Therefore, we have: . $\frac{1}{2}S\:=\:2\quad\Rightarrow\quad \boxed{S\,=\,4}$

• Jul 9th 2006, 10:41 AM
Nichelle14
It is the third choice. n-1 is an exponent of 2.

Thanks
• Jul 9th 2006, 07:14 PM
malaygoel
Quote:

Originally Posted by rgep
Here's a general technique that may help. I'll tackle the example $\frac{n}{2^{n-1}}$. The trick is to write down a function $f(x)$ for which the sum is a special value. Here we consider $f(x) = \sum_n \frac{n}{x^{n-1}}$, so that you want $f(2)$. We look at $x^{-2} f(x) = \sum_n \frac{n}{x^{n+1}}$ in order to be able to integrate term by term. So $x^{-2}f(x)$ is the derivative of $\sum_n \frac{-1}{x^n}$, which is $\frac{-1}{1-1/x} = \frac{x}{1-x} = 1+\frac{1}{1-x}$. Hence $x^{-2} f(x) = \frac{1}{(1-x)^2}$ and your sum is $f(2) = 4$.

Rgep, Will you please explain the use of integration and derivative here. I am not able to get it.

Keep Smiling
Malay
• Jul 9th 2006, 09:17 PM
CaptainBlack
Quote:

Originally Posted by malaygoel
Rgep, Will you please explain the use of integration and derivative here. I am not able to get it.

Keep Smiling
Malay

When you have a series of the form:

$
S(c,a,b)=\sum_{n=a}^b n.c^{n-1}
$

with $a,\ b,\$ and $c$ numerical constants,
we consider the function:

$
S(x,a,b)=\sum_{n=a}^b n.x^{n-1}
$

each term of this is obviously the derivative wrt $x$of:

$
t(x,n)=x^n
$
.

So we can integrate term by term to get:

$
I(x,a,b)=\sum_{n=a}^b x^{n}
$

Now if RHS is a geometric series and may be summed:

$I(x,a,b)=x^a \frac{1-x^{b-a+1}}{1-x}$,

(you will need to check this last sum, I did it in a bit of a rush :( )

and:

$
S(c,a,b)=\left{} \frac{d}{dx}I(x,a,b) \right|_{x=c}
$

Now you need only play around with the values of a, b and c (and the odd
limiting process to get the required result).

This is quite a common trick employed to find the sum of series.

RonL