hello I am given prove that

what I have done

this is a Geometric progression therefore sum to infinity =

a=

r =

r =

substituting back into the equation I get

sum to infinity =

can someone please show me where I am going wrong, thanks.

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- April 9th 2013, 08:55 PMsigma1series proof
hello I am given prove that

what I have done

this is a Geometric progression therefore sum to infinity =

a=

r =

r =

substituting back into the equation I get

sum to infinity =

can someone please show me where I am going wrong, thanks. - April 10th 2013, 12:12 AMagentmulderRe: series proof
- April 10th 2013, 12:25 AMsigma1Re: series proof
- April 10th 2013, 01:21 AMagentmulderRe: series proof
Your problem requires cleverness i do not have at this time so i am looking forward to a solution posted by another MHF member. The series is interesting.

:)

WIA is also stumped.

series(sin(nx))/(2^n&# 41; , n = 1 to infinity) - Wolfram|Alpha - April 10th 2013, 01:23 AMProve ItRe: series proof
- April 10th 2013, 08:18 AMGusbobRe: series proof
I'll replace with for convenience.

Using the identity , your series may be rewritten as

.

Individually, these two sums are convergent geometric series (do you know why?). Calculate the sum for each term and add them up. I have verified that you do indeed get the desired result (after some algebraic manipulations), but I have no desire to type it out. You should be able to work it out yourself.

You may need to justify why you are able to break the sum into two infinite series, but this is a straightforward application of limit theorems since both your series converge.