# Confusion about radius of convergence of a power series

I'm a bit confused about the following statement from a script. We look at the power series with coefficient $$a_k = (1+1/k)^{k^2}$$ and want to compute the radius of convergence $$R$$ of the series. The solution given is $$R=1$$, applying the $$k^2$$ root to $$a_k$$. I don't get this: The Cauchy-Hadamard formula involves the $$k$$th root which would give $$R=1/e$$ in my opinion. Is this a mistake in the script or is it me who is mistaken?

• You are correct; it's likely a typo. – T. Bongers Dec 7 '18 at 1:16
• You do not say what power series actually is. If it is $$\sum (1+1/k)^{k^2}\;x^{k^2}$$ then you should use the $k^2$ root. – GEdgar Dec 7 '18 at 12:16

## 2 Answers

You are indeed correct.

For coefficients $$a_k:=(1+\frac {1}{k})^{k^2}$$, the Cauchy-Hadamard formula gives:

$$R=\frac {1}{\limsup_{k\to\infty}(|a_k|)^{\frac {1}{k}}}=\frac {1}{\limsup_{k\to\infty}(|(1+\frac {1}{k})^{k^2}|)^{\frac {1}{k}}}=\frac {1}{\limsup_{k\to\infty}((1+\frac {1}{k})^{k^2})^{\frac {1}{k}}}=\frac {1}{\limsup_{k\to\infty}(1+\frac {1}{k})^k}=\frac {1}{e}$$

Just another way to do it. $$a_k=(1+\frac {1}{k})^{k^2}\implies \log(a_k)=k^2\log \left(1+\frac{1}{k}\right)$$ $$\log(a_k)-\log(a_{k+1})=k^2\log \left(1+\frac{1}{k}\right)-(k+1)^2\log \left(1+\frac{1}{k+1}\right)$$ Now, use Taylor expansions to get $$\log(a_k)-\log(a_{k+1})=-1+\frac{1}{3 k^2}+O\left(\frac{1}{k^3}\right)$$ Continue with Taylor $$\frac{a_k} {a_{k+1}}=\frac{1}{e}+\frac{1}{3 e k^2}+O\left(\frac{1}{k^3}\right)$$ making $$R=\frac{1}{e}$$ as you found.