i have an exam and I'm stuck on this problem dealing with power series expansion. Can someone please help?

Question

problem:
consider the function y(x), given by
y(x)=summation (0 to infiniti) [x^(2n)]/(1-2n)

find a power series expansion of (integral symbol) y(x) dx

so far i have...
y(x) = ∑ [x^(2n)] / (1-2n)...........n=0 to infinity

y(x) = 1 + [ (x^2) / (-1) ] + [ (x^4) / (-3) ] + [ (x^6) / (-5) ] + ...

= 1 - (x^2) - (1/3)*(x^4) - (1/5)*(x^6) - . . .

∫ y(x) dx = x - [ (x^3) / 3 ] - [ (x^5) / (3*5) ] - [ (x^7) / (5*7) ] - . . .

my problem is that i can't find the shorthand sigma notation for this result and i don't know how. it's really bothering me. your help is much appreciated. thank you

Answer 1

Don't rewrite the sum explicitly. Integrate [x^(2n)] / (1 - 2n) "symbolically", i.e., as

[x^(2n + 1)] / [(2n + 1)(1 - 2n)]

so

∫ y(x) dx = ∑ [x^(2n + 1)] / [(2n + 1)(1 - 2n)] ....... n=0, ∞

which is equivalent to

∫ y(x) dx = -∑ [x^(2n + 1)] / (4n² - 1) ....... n=0, ∞