멱급수와 수렴 반경

We'll start by applying the ratio test twice,

• once for the series whose $n^{\rm th}$ term is $a_n=\frac{x^n}{3^{2n}\log n}$, and
• once for the series whose $n^{\rm th}$ term is $a_n=\big|\frac{x^n}{3^{2n}\log n}\big|$.

For both series, i.e. for both $a_n$'s,

$$\begin{aligned} \lim_{n\to\infty} \bigg| \frac{ a_{n+1} }{ a_n } \bigg| &= \lim_{n\to\infty} \bigg| \frac{x^{n+1}}{3^{2(n+1)}\log(n+1)} \ \frac{3^{2n}\log n}{x^n} \bigg| \\ &= \lim_{n\to\infty} \bigg| \frac{x\log n}{3^2\log(n+1)} \bigg| =\lim_{n\to\infty} \bigg| \frac{x\log n}{3^2[\log(n)+\log(1+1/n)]} \bigg| \\ &=\lim_{n\to\infty} \bigg| \frac{x}{3^2[1+\log(1+1/n)/\log(n)]} \bigg| \\ &=\frac{|x|}{9} \end{aligned}$$

Therefore, by the ratio test, when $|x|<9$, the series $\sum_{n=2}^\infty \big|\frac{x^n}{3^{2n}\log n}\big|$ converges and the series $\sum_{n=2}^\infty \frac{x^n}{3^{2n}\log n}$ converges absolutely, and when $|x|> 9$ the series $\sum_{n=2}^\infty \frac{x^n}{3^{2n}\log n}$ diverges. That just leaves $x=\pm 9$.

For $x=-9$, $\displaystyle\sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} =\sum_{n=2}^\infty\frac{(-1)^n}{\log n}$ which converges by the alternating series test.

For $x=+9$, $\displaystyle\sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} =\sum_{n=2}^\infty\frac{1}{\log n}$ which is the same series as $\displaystyle\sum_{n=2}^\infty\Big|\frac{(-1)^n}{\log n}\Big|$. We shall shortly show that $n\ge\log n$, and hence $\frac{1}{\log n}\ge \frac{1}{n}$ for all $n\ge 1$. This implies that the series $\displaystyle\sum_{n=2}^\infty\frac{1}{\log n}$ diverges by comparison with the divergent series $\displaystyle\sum_{n=2}^\infty\frac{1}{n^p}\bigg|_{p=1}$. This yelds both divergence for $x=9$ and also the failure of absolute convergence for $x=-9$.

Finally, we show that $n-\log n > 0$, for all $n\ge 1$. Set $f(x)=x-\log x$. Then $f(1)=1>0$ and

$$f'(x) = 1 -\frac{1}{x} \ge 0\qquad\text{for all }x\ge 1$$

So $f(x)$ is (strictly) positive when $x=1$ and is increasing for all $x\ge 1$. So $f(x)$ is (strictly) positive for all $x\ge 1$.

We recognize $\displaystyle \sum_{n=3}^\infty \frac{n}{5^{n-1}}$ as $f(x)=\displaystyle \sum_{n=3}^\infty n\cdot x^{n-1}$, evaluated at $x=\dfrac15$. We should figure out what $f(x)$ is in equation form (as opposed to power series form). Notice that this looks similar to the derivative of the geometric series $\displaystyle\sum x^n$.

$$\begin{aligned} \frac{1}{1-x}&=\sum_{n=0}^\infty x^n \qquad \text{when $|x|<1$}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{1-x}\right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sum_{n=0}^\infty x^n\right\}\\ \frac{1}{(1-x)^2}&=\sum_{n=1}^\infty nx^{n-1}\\ &= 1x^0 + 2x^1+\sum_{n=3}^\infty nx^{n-1}\\ &= 1 + 2x+\sum_{n=3}^\infty nx^{n-1}\\ \text{So,}\qquad \frac{1}{(1-x)^2}-1-2x&=\sum_{n=3}^\infty nx^{n-1}\\ \text{Setting $x=\dfrac15$:}\qquad \frac{1}{(1-1/5)^2}-1-\frac25&=\sum_{n=3}^\infty n\left(\frac15\right)^{n-1}\\ \left(\frac{5}{4} \right)^2-1-\frac25&=\sum_{n=3}^\infty \frac{n}{5^{n-1}}\\ \end{aligned}$$

So, our series evaluates to $\displaystyle\frac{25}{16}-1-\frac{2}{5}=\frac{13}{80}$.