$\sum_{i=0}^{\infty} e^{-\sqrt n}z^n $

I tried to find the radius of convergence of a power series.. is this equation a geometric series?

or would it be easier to do a ratio test and

$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|=L$ Then radius of convergence $R=1/L$?

I'm got it converges to $0$, not confident about it though.

  • $\begingroup$ Cauchy-Hadamard Radius Formula. $\endgroup$ Commented Oct 11, 2018 at 17:52

1 Answer 1


But you can use that ratio test:\begin{align}\lim_{n\to\infty}\frac{e^{-\sqrt{n+1}}}{e^{-\sqrt n}}&=\lim_{n\to\infty}e^{\sqrt n-\sqrt{n+1}}\\&=e^{\lim_{n\to\infty}\sqrt n-\sqrt{n+1}}\\&=e^0\\&=1.\end{align}Therefore, the radius of convergence is $1$.

  • $\begingroup$ How do you do line 2 and 3? I'm quite lost on that :( $\endgroup$ Commented Oct 11, 2018 at 17:31
  • $\begingroup$ Do you mean the second and the third $=$ signs? $\endgroup$ Commented Oct 11, 2018 at 17:33
  • $\begingroup$ where you took e out of the limit and made it 0.. I'm curious of the steps $\endgroup$ Commented Oct 11, 2018 at 19:07
  • 1
    $\begingroup$ @MathsGoogle Since the exponential map is continuous, whenever you have a convergen sequence $(a_n)_{n\in\mathbb N}$, you have$$\lim_{n\to\infty} e^{a_n}=e^{\lim_{n\to\infty}a_n}.$$And\begin{align}\lim_{n\to\infty}\sqrt n-\sqrt{n+1}&=\lim_{n\to\infty}\frac{\left(\sqrt n-\sqrt{n+1}\right)\left(\sqrt n+\sqrt{n+1}\right)}{\sqrt n+\sqrt{n+1}}\\&=\lim_{n\to\infty}\frac{-1}{\sqrt n+\sqrt{n+1}}\\&=0.\end{align} $\endgroup$ Commented Oct 11, 2018 at 20:19

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