How to calculate the sum?

Question

infinity
SUM (1/3^n)*sin(((2n+1)pi)/2)
n=1

Thanks :)

Answer 1

sin(((2n+1)pi)/2)=-1if n is odd
sin(((2n+1)pi)/2)=1if n is even
so u get sum as
-1/3+1/3^2-1/3^3+...............
which is a GP
=(-1/3)/[1-(-1/3)]
=(-1/3)/{1+1/3}
=(-1/3)/[4/3}
=-1/4

Answer 2

ew know

e^ix = cos x + i sin x

so te^(ix) = t(cos x + i sin x)

so e^t sin x is imaginary part of e^(t+ix)
now (1/3^n)*sin(2n+1)pi/2) is imaginary part of

(1/3^n) e^(2n+1)i pi/2
take the above and find the sum and take the imaginary part you have the result.
Above is a GP