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,
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 $R=1/L$.