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:

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 . 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 .

- Jul 9th 2006, 04:59 AMgalactus
I'll try the

You can just use the closed form for a geometric series, namely,

Subtract your 1 and you have 1 as your answer.

can be treated similarly.

This is the same as:

- Jul 9th 2006, 05:26 AMSoroban
Hello, Nichelle14!

I'll assume that the exponent is .

Quote:

Evaluate: .

Here's an elementary approach . . .

We are given: .

Divide by 2: .

Subtract:. . . .

The right side is a geometric series with

. . Hence, its sum is: .

Therefore, we have: .

- 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**

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