What is the value of the infinite series?

[ptolema]

Homework Statement

evaluate the sum:

(it may help to think of this as a value of a function defined by a power series.)

Homework Equations

as a function defined by a power series, the function is centered at 0: f(1/3)=

=1/3+4/9+1/3+16/81+...+n²/3ⁿ

The Attempt at a Solution

i'm not entirely clear on how to actually evaluate the value of an infinite series. i could definitely use a few pointers

 

[tiny-tim]

hi ptolema!

(have a sigma: ∑ )

(not sure about the power series, but … )

have you tried ∑ (n+1)²/3ⁿ - ∑ n²/3ⁿ ?

 

[lanedance]

ok so how about considering a taylor series about zero as a start
f(a) = \sum_{n=0} f^n(a) x^n

then re-write your series as
0 + 1.\frac{1}{3} + 2^2.\frac{1}{3^2} +3^2.\frac{1}{3^3} +...+ n^2 (\frac{1}{3^n})

 

[Gib Z]

\sum_{n=1}^{\infty} \frac{x^n}{3^n} = \frac{x}{3-x} for |x| < 3.

Try fiddling around with various manipulations such as differentiating both sides, and try to lead yourself to a series that will drop out your required sum after a suitable substitution for x.

 

[dextercioby]

Going along the hint by tiny-tim, complete the dots below

\sum_{n=1}^{\infty} \frac{n^2}{3^n} = \sum_{n=0}^{\infty} \frac{(n+1)^2}{3^{n+1}} = \frac{1}{3}\sum_{n=0}^{\infty} \frac{n^2 + 2n +1}{3^n} =...