How is the Radius of Convergence of a Series determined?

Question

Consider $$\sum_{n=0}^{\infty}\frac{(-1)^nx^n}{(n+1)^2}$$ which by the ratio test the ratio of two consecutive terms converges to $|x|$ as $n\rightarrow \infty$ and has a radius of convergence equal to $1$.

Now consider $$\sum_{n=0}^{\infty}(-1)^n(2^n+n^2)x^n$$ which by the ratio test the ratio of two consecutive terms converges to $2|x|$ as $n\rightarrow \infty$ and has a radius of convergence equal to $\frac{1}{2}$.

My question is why the radius of convergence takes these values ($1$ for the former and $\frac{1}{2}$ for the latter)?

You have my sympathy if this is blatantly obvious to you, but it is not clear to me. So could someone please explain in simple English why the radius of convergence takes those values above?

Or, put in another way, from the second summation you know by the ratio test that the ratio of two of its consecutive terms converges to $2|x|$ as $n\rightarrow \infty$ how then do you proceed to determine the radius of convergence?

Many thanks.

Answer 1

The formula for the Radius of Convergence of $$ \sum_{n=0}^\infty a_nx^n $$ is $$ R=\left(\limsup_{n\to\infty}\left|a_n\right|^{1/n}\right)^{-1} $$ This formula is derived using the Ratio Test.

It is pretty easy to apply this to the series in the question.

$$ \begin{align} \lim_{n\to\infty}\left(\frac1{(n+1)^2}\right)^{1/n} &=\lim_{n\to\infty}(n+1)^{-2/n}\\ &=\left(\lim_{n\to\infty}n^{1/n}\right)^{-2}\lim_{n\to\infty}\left(1+\frac1n\right)^{-2/n}\\[9pt] &=1^{-2}\cdot1^0\\[12pt] &=1 \end{align} $$ giving a radius of convergence of $1$.

Likewise, $$ \begin{align} \lim_{n\to\infty}\left(2^n+n^2\right)^{1/n} &=2\lim_{n\to\infty}\left(1+\frac{n^2}{2^n}\right)^{1/n}\\ &=2\cdot1^0\\[6pt] &=2 \end{align} $$ giving a radius of convergence of $\frac12$.

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As you say, in the second series, the ratio of the absolute values of the terms tends to $2|x|$. For the series to converge, the ratio needs to be $\le1$. Thus, we need $|x|\le\frac12$. Therefore, the radius of convergence is $\frac12$.

Answer 2

An interesting aspect is to evaluate the series of the question. Consider \begin{align} S_{1}(x) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, x^{n}}{(n+1)^{2}} \\ S_{2}(x) &= \sum_{n=0}^{\infty} (-1)^{n} \, (2^{n} + n^{2}) \, x^{n}. \end{align} The first series: \begin{align} \partial_{x} \left(x \, S_{1}(x) \right) &= \sum_{n \geq 0} \frac{(-1)^{n} \, x^{n}}{n+1} = \sum_{n \geq 1} \frac{(-1)^{n-1} \, x^{n-1}}{n} = \frac{\ln(1+x)}{x}. \end{align} or $$S_{1}(x) = - \frac{1}{x} \, Li_{2}(-x),$$ where $Li_{2}(x)$ is the dilogarithm function. The second series follows from $$\sum_{n=0}^{\infty} n^{2} \, x^{n} = \frac{x(1+x)}{(1-x)^{3}}$$ and is seen to be $$S_{2}(x) = \frac{1}{1+2x} - \frac{x(1-x)}{(1+x)^{3}}.$$