- #1

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## Homework Statement

[tex]

\sum_{n=1}^{\infty}\frac{sinn}{2^n}

[/tex]

## Homework Equations

Definition of a geometric series:

[tex]

\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}

[/tex]

## The Attempt at a Solution

Basically I can use the geometric series idea and implement it into the denominator of the question (i.e. sub x=2 into the equation from part b and change the lower index to n=1)

[tex]

\sum_{n=1}^{\infty}2^n=\frac{1}{1-2}

[/tex]

Taking the derivate of both sides:

[tex]

\sum_{n=1}^{\infty}n2^{n-1}=\frac{1}{(1-2)^2}

[/tex]

Multiplying both sides by 2:

[tex]

\sum_{n=1}^{\infty}n2^n=\frac{2}{(1-2)^2}

[/tex]

Of course I can simplify the question furthermore and get a single value as a result, my problem is I don't understand on how to incorporate the numerator (sin n) into the problem. Is my approach correct, or am I substituting the wrong x-value?

Thanks in advance for any help/advice. :)