#### Question

# How to find the radius of convergence?

How do you find the radius of convergence for the sum from 0 to +infinity of (2^n + n^2)x^n ?

#### Answers

Using the Ratio Test:

r = lim(n→∞) |[2^(n+1) + (n+1)^2] x^(n+1) / [(2^n + n^2) x^n]|

..= |x| * lim(n→∞) [2 * 2^n + (n+1)^2] / (2^n + n^2)

..= |x| * lim(n→∞) [2 * 2^n ln 2 + 2(n+1)] / (2^n ln 2 + 2n), by L'Hopital's Rule

..= |x| * lim(n→∞) [2 * 2^n (ln 2)^2 + 2] / (2^n (ln 2)^2 + 2), by L'Hopital's Rule

..= |x| * lim(n→∞) 2 * 2^n (ln 2)^3 / (2^n (ln 2)^3), by L'Hopital's Rule

..= 2 |x|.

So, this series converges (save a possible endpoint or two) when

r = 2|x| < 2 ==> |x| < 1/2.

Hence, the radius of convergence is 1/2.

I hope this helps!

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