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,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
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 $\displaystyle f(2) = 4$.
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:
$\displaystyle \frac{n}{2^{n}2^{-1}}$
$\displaystyle 2\sum_{n=1}^{\infty}\frac{n}{2^{n}}=\frac{2}{1-\frac{1}{2}}=4$
Hello, Nichelle14!
I'll assume that the exponent is $\displaystyle n-1$.
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: .$\displaystyle \frac{1}{2}S\:=\:2\quad\Rightarrow\quad \boxed{S\,=\,4}$
When you have a series of the form: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