Find the radius of convergence and the interval of convergence of the power series


My attempt: The root test brings us to


but I'm not sure where this brings me. Is the ratio test any help?

Any help appreciated!

  • 1
    $\begingroup$ What is the limit for $e$? $\endgroup$ – Ahmed S. Attaalla Feb 11 '17 at 17:23
  • 1
    $\begingroup$ $\lim (1+1/n)^n$ is one of the best known limits in the world. $\endgroup$ – zhw. Feb 11 '17 at 17:26

Note: Both tests, root test as well as ratio test can be used to analyze the convergence behavior. But, the first one is in this case considerably simpler to use.

We recall the validity of the limit \begin{align*} \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e \end{align*}

Root test:

We obtain \begin{align*} \lim_{n\rightarrow\infty}\left|\sqrt[n]{\left(1+\frac{1}{n}\right)^{n^2}x^n}\right| =|x|\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^{n} =|x|e \end{align*}

and we conclude the series is convergent for \begin{align*} |x|<\frac{1}{e} \end{align*} and divergent for $|x|>\frac{1}{e}$.

$$ $$

Ratio test:

We obtain \begin{align*} \lim_{n\rightarrow\infty}\left|\frac{\left(1+\frac{1}{n+1}\right)^{(n+1)^2}x^{n+1}} {\left(1+\frac{1}{n}\right)^{n^2}x^n}\right| &=|x| \lim_{n\rightarrow\infty}\frac{\left(1+\frac{1}{n+1}\right)^{n^2+2n+1}} {\left(1+\frac{1}{n}\right)^{n^2}}\\ &=|x|\lim_{n\rightarrow\infty}\left(\frac{n(n+2)}{(n+1)(n+1)}\right)^{n^2} \left(1+\frac{1}{n+1}\right)^{2n+1}\\ &=|x|\lim_{n\rightarrow\infty} \underbrace{\frac{1}{\left(1+\frac{1}{n^2+2n}\right)^{n^2}}}_{\longrightarrow \ \ \frac{1}{e}} \underbrace{\left(1+\frac{1}{n+1}\right)^{2n}}_{\longrightarrow\ \ e^2} \underbrace{\left(1+\frac{1}{n+1}\right)^{1}}_{\longrightarrow \ \ 1}\\ &=|x|e \end{align*} and we conclude as above.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.