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?
Here's a general technique that may help. I'll tackle the example . The trick is to write down a function for which the sum is a special value. Here we consider , so that you want . We look at in order to be able to integrate term by term. So is the derivative of , which is . Hence and your sum is .
When you have a series of the form:Originally Posted by malaygoel
with and numerical constants,
we consider the function:
each term of this is obviously the derivative wrt of:
.
So we can integrate term by term to get:
Now if RHS is a geometric series and may be summed:
,
(you will need to check this last sum, I did it in a bit of a rush )
and:
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