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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,

jack
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  • Who is $L$? We need more context. Edit your answer and add the details. – Carlos Jiménez Mar 19 '18 at 00:11
  • Do you know the expression for the radius of convergence of $\sum_k c_k x^k$ ? – Prasun Biswas Mar 19 '18 at 00:18
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    You can compute the radius of convergen using this formula. It involves computing $\lim_{k\to\infty}\left|\frac{a_k}{k+1}\right|^{1/k}=\lim_{k\to\infty}|a_k|^{1/k}/\lim_{k\to\infty}(k+1)^{1/k}$. Use that $\lim_{k\to\infty}(k+1)^{1/k}=1$ and the limit you were given. – blueInk Mar 19 '18 at 00:28
  • Thank you for you comment so I'm supposed to use the root test, but I'm still not sure how to compute limk→∞|ak|^1/k, I know from the question that it is >0 but I don't see how to get to the radius of convergence.@blueInk – jack Mar 19 '18 at 01:41
  • If I just sub limk→∞|ak|^1/k=L and limk→∞(k+1)1/k=1, doesn't that give me R=L/1 not R=1/L? @blueInk – jack Mar 19 '18 at 01:50
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    But look at the formula in the link. After computing the limit I wrote one must take reciprocal to get the radius. – blueInk Mar 19 '18 at 01:53

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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 $R=1/L$.

mathematics2x2life
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