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?

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- Jul 8th 2006, 05:48 PMNichelle14Evaluate 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 PMearbothQuote:

Originally Posted by**Nichelle14**

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

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

So sorry I can't help you.

Greetings

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

- Jul 9th 2006, 04:59 AMgalactus
I'll try the $\displaystyle \sum_{n=1}^{\infty}\frac{n}{2^{n}}-1$

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

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

Subtract your 1 and you have 1 as your answer.

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

This is the same as:

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

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

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

Quote:

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

Here's an elementary approach . . .

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

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

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

The right side is a geometric series with $\displaystyle a = 1,\;r = \frac{1}{2}$

. . Hence, its sum is: .$\displaystyle \frac{1}{1 - \frac{1}{2}} \,= \,2$

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

- Jul 9th 2006, 10:41 AMNichelle14
It is the third choice. n-1 is an exponent of 2.

Thanks - Jul 9th 2006, 07:14 PMmalaygoelQuote:

Originally Posted by**rgep**

Keep Smiling

Malay - Jul 9th 2006, 09:17 PMCaptainBlackQuote:

Originally Posted by**malaygoel**

$\displaystyle

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

$

with $\displaystyle a,\ b,\ $ and $\displaystyle c$ numerical constants,

we consider the**function**:

$\displaystyle

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

$

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

$\displaystyle

t(x,n)=x^n

$.

So we can integrate term by term to get:

$\displaystyle

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

$

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

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

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

and:

$\displaystyle

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