Show that sum (1 -> inf) k^2/2^k converges and find the sum

Show that sum (1 -> inf) k^2/2^k converges (using the root and ratio tests) and find the sum using power series.


Thank you for your help!!! :)

Using the Ratio Test:
r = lim(k→∞) [(k+1)^2 / 2^(k+1)] / [k^2 / 2^k]
..= lim(k→∞) (k+1)^2 / (2k^2)
..= 1/2, by comparing leading coefficients (or L'Hopital's Rule twice)

Since r = 1/2 < 1, this series converges.
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From your previous post, we found that
x(1 + x)/(1 - x)^3 = Σ(k = 1 to ∞) k^2 x^k, valid for |x| < 1.

Let x = 1/2:
Σ(k = 1 to ∞) k^2 (1/2)^k = (1/2)(1 + 1/2)/(1 - 1/2)^3
==> Σ(k = 1 to ∞) k^2 / 2^k = 6.

I hope this helps!