Find the `underset(k=1)overset(oo)sumunderset(n=1)overset(oo)sumk/(2^(n+k))`.

underset(n=1)overset(oo)sum(x^(n))/(n+2)=

underset(n=1)overset(oo)sum(n^(2))/((n+1)!)=

underset(n=1)overset(oo)Sigma(1)/((2n-1)!)=

underset(n=1)overset(oo)sum(x^(n+1//2))/(n+1)=

If x = underset(n-0)overset(oo)sum a^(n), y= underset(n =0)overset(oo)sum b^(n), z = underset(n =0)overset(oo)sum C^(n) where a,b,c are in A.P. and |a| lt 1, |b| lt 1, |c| lt 1 , then x,y,z are in

underset(n=0)overset(oo)Sigma(1)/((n+1)!)=

For 0ltphilepi//2 , if : x=underset(n=0)overset(oo)sumcos^(2n)phi,y=underset(n=0)overset(oo)sumsin^(2n)phi , z=underset(n=0)overset(oo)sumcos^(2n)sin^(2n)phi , then :

underset(n=1)overset(oo)Sigma(2n)/((2n+1)!)=