Radius of convergence power series. (Homework help)

Question

Suppose $\sum_{k=0}^{\infty} {a_k}{x^k}$ is a power series and $$\lim_{k\rightarrow\infty}|a_k|^\dfrac{1}{k}$$=L>0 converges

Prove that $\sum_{k=0}^{\infty} \dfrac{a_k}{k+1}{x^k}$ Has a radius of convergence R=$\dfrac{1}{L}$.

I have no idea where to start, all of the examples I have seen use the ratio test to find the radius of convergence, (usually something like $\sum_{k=0}^{\infty} \dfrac{x^n}{n!}$ or $\sum_{k=0}^{\infty} \dfrac{x^2n}{n(2n)}$ I can find the radius of convergence for these). But I haven't seen any examples with ${a_k}$ in the series. Any help would be greatly appreciated. Thank you,

Answer 1

The general formula for the radius of convergence comes straight from the Root Test, as in blueInk's comments. To keep the algebra simple, I shall simply apply the Root Test itself: $$ \lim_{k \to \infty} \left|\dfrac{a_k}{k+1} x^k\right|^{1/k}= |x| \lim_{k \to \infty} \left|\dfrac{a_k}{k+1}\right| $$ But $\lim_{k \to \infty} |a_k|^{1/k}=L$ and $\lim_{k \to \infty} |k+1|^{1/k}=1$ so we have $$ \lim_{k \to \infty} \left|\dfrac{a_k}{k+1} x^k\right|^{1/k}= |x| \dfrac{\lim_{k \to \infty}|a_k|^{1/k}}{\lim_{k \to \infty} (k+1)^{1/k}}= L|x| $$ But this converges for $L|x|<1$ so that $|x|<1/L$. Then the radius of convergence is $R=1/L$.