# Radius of convergence of $\sum_{n} \frac{x^n}{n\sqrt{n}}$

Trying to find the radius of convergence for $\displaystyle \sum_{n} \frac{x^n}{n\sqrt{n}}$

I apply the root test: $\displaystyle \lim_{n \to \infty} \left(\bigg|\frac{x^n}{n\sqrt{n}}\bigg|\right)^{1/n} =\lim_{n \to \infty} \frac{|x|}{n^{1/n}n^{1/2n}} = |x|.$

The series converges absolutely if $|x| < 1$ therefore $R = 1$.

Is the above correct? On Wikipedia I find the formula:

$R = 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}$

How does one use this to find $R$ in the above case?

## 5 Answers

The formula $R = 1/\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}}$ is for the series $\displaystyle\sum_{n = 0}^{\infty}c_nx^n$.

Note that $c_n$ is the coefficient of the $x^n$ term, not the $n$-th term itself.

So, here $c_n = \dfrac{1}{n\sqrt{n}} = n^{-3/2}$, and we need to compute $\displaystyle\limsup_{n \rightarrow \infty}{\sqrt[n]{|c_n|}} = \displaystyle\limsup_{n \rightarrow \infty}n^{-\tfrac{3}{2n}}$.

By taking the logarithm and using L'Hoptital's Rule (or whatever other method you prefer) you get $\displaystyle\lim_{n \rightarrow \infty}n^{-\tfrac{3}{2n}} = 1$.

Since $\displaystyle\lim_{n \rightarrow \infty}n^{-\tfrac{3}{2n}}$ exists, we have $\displaystyle\limsup_{n \rightarrow \infty}n^{-\tfrac{3}{2n}} = \displaystyle\lim_{n \rightarrow \infty}n^{-\tfrac{3}{2n}} = 1$, and thus, $R = 1$.

The answer is correct, and the two formulas are equivalent if $$\lim_ \limits{n \to \infty}|c_n|^{1/n}$$exists.

The ratio test is however the easier test in this case. Observe

$$\lim_ \limits{n \to \infty} \left| \frac{x^{n+1}}{(n+1)^{3/2}} \frac{n^{3/2}}{x^n}\right| = |x|,$$ and since ratio test implies absolute convergence when the ratio is less than $1$, we get a radius of convergence of $1$.

• I think you want $n \to \infty$ and not $x \to \infty$. – Jendrik Stelzner Dec 30 '15 at 23:59
• @JendrikStelzner You are very correct. Thanks! – GaussTheBauss Dec 31 '15 at 0:00

First you have to calculate $$\limsup_{n \to \infty} \left( \frac{1}{n \sqrt{n}} \right)^{1/n} = \limsup_{n \to \infty} \left( \frac{1}{n^{3/2}} \right)^{1/n} = \limsup_{n \to \infty} \left( \frac{1}{n^{1/n}} \right)^{3/2}.$$ Because $\lim_{n \to \infty} n^{1/n} = 1$ it follows that $$\limsup_{n \to \infty} \left( \frac{1}{n^{1/n}} \right)^{3/2} = \lim_{n \to \infty} \left( \frac{1}{n^{1/n}} \right)^{3/2} = \left( \lim_{n \to \infty} \frac{1}{n^{1/n}} \right)^{3/2} = 1$$ because the limit exists. The radius of convergence is now given by $$\frac{1}{\limsup_{n \to \infty} \left( \frac{1}{n \sqrt{n}} \right)^{1/n}} = \frac{1}{1} = 1.$$

What you wrote is correct, from the root test the radius of convergence is $R=1$.

You may observe that both series $\displaystyle \sum_{n} \frac{(-1)^n}{n\sqrt{n}}$ and $\displaystyle \sum_{n} \frac{1}{n\sqrt{n}}$ are also absolutely convergent: $$\left|\sum_{n\geq1} \frac{(-1)^n}{n\sqrt{n}}\right|\leq \sum_{n\geq1}\left| \frac{1}{n\sqrt{n}}\right|=\sum_{n\geq1}\frac1{n^{3/2}}<\infty$$ as a $p$-series with $p>1$.

$$\lim_{n\to\infty} \left|x^nn^{-\dfrac32}\right|^{\dfrac1n} = |x|\lim_{n\to\infty} \exp\left(-\dfrac 32\dfrac {\log n}n\right) = |x|\exp(0) = |x|.$$