13. 멱급수Power Series

Proof. For $n=0,1,2,\ldots$, let $\displaystyle s_n(x) = \sum_{k=0}^{n} x^k = 1+x+x^2+\cdots+x^n$ denote the $n$th partial sum.

For $x\neq 1$, $(1-x)s_n(x) = (1-x)(1+x+\cdots+x^n) = 1-x^{n+1}$, so $$s_n(x) = \frac{1-x^{n+1}}{1-x}.$$

If $-1 \lt x \lt 1$, then $|x| \lt 1$, so $x^{n+1}\to 0$ as $n\to\infty$. Hence $$f(x) = \lim_{n\to\infty} s_n(x) = \lim_{n\to\infty}\frac{1-x^{n+1}}{1-x} = \frac{1-0}{1-x} = \frac{1}{1-x} = g(x).$$ That is, $f(x)=g(x)$ on $-1 \lt x \lt 1$.

Solution. The given series is $$f(x) = \sum_{n=0}^{\infty} \left(-\frac12\right)^n (x-2)^n = \sum_{n=0}^{\infty} \left(-\frac{x-2}{2}\right)^n,$$ a geometric series with first term $1$ and common ratio $\displaystyle r = -\frac{x-2}{2}$.

A geometric series converges when $|r| \lt 1$, so $$\left|\frac{x-2}{2}\right| \lt 1 \iff |x-2| \lt 2 \iff 0 \lt x \lt 4,$$ and for such $x$, $f(x)$ converges.

Its sum is $$f(x) = \frac{1}{1-r} = \frac{1}{1+\dfrac{x-2}{2}} = \frac{2}{2+(x-2)} = \frac{2}{x}.$$

Solution. To expand around $x=1$, rewrite $2/x$ as $2/(1+(x-1))$.

For $|x-1| \lt 1$, substituting $u=-(x-1)$ into the geometric series formula $\displaystyle \sum_{n=0}^{\infty}u^n=\frac{1}{1-u}$ ($|u| \lt 1$) gives $$\frac{1}{1+(x-1)} = \sum_{n=0}^{\infty}(-(x-1))^n = \sum_{n=0}^{\infty}(-1)^n(x-1)^n,$$ so $$f(x)=\frac{2}{x} = \frac{2}{1+(x-1)} = \sum_{n=0}^{\infty}2(-1)^n(x-1)^n.$$

This series converges for $|x-1| \lt 1$ and diverges for $|x-1|\ge 1$, so its radius of convergence is $R=1$.

Solution.

(1) Let $\displaystyle a_n = \frac{(-1)^{n-1}}{n}$. By the Ratio Test, $$\lim_{n\to\infty}\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n\to\infty}\frac{n}{n+1}|x| = |x|,$$ so the series converges absolutely for $|x| \lt 1$ and diverges for $|x| \gt 1$. The radius of convergence is $R=1$.

At $x=1$, the series is $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}$, which converges by the Alternating Series Test. At $x=-1$, the series is $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}(-1)^n}{n}=\sum_{n=1}^{\infty}\left(-\frac1n\right)$, which is $-1$ times the harmonic series and hence diverges. Thus the interval of convergence is $(-1,1]$.

(2) Let $\displaystyle a_n=\frac{(-1)^{n-1}}{2n-1}$. Then $$\lim_{n\to\infty}\left|\frac{a_{n+1}x^{2(n+1)-1}}{a_n x^{2n-1}}\right| = \lim_{n\to\infty}\frac{2n-1}{2n+1}|x|^2 = |x|^2,$$ so the series converges absolutely for $|x|^2 \lt 1$, i.e. $|x| \lt 1$, and diverges for $|x| \gt 1$. The radius of convergence is $R=1$.

At $x=1$, the series is $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}$, which converges by the Alternating Series Test. At $x=-1$, since $x^{2n-1}=(-1)^{2n-1}=-1$, the series is $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cdot(-1)}{2n-1}=\sum_{n=1}^{\infty}\frac{(-1)^n}{2n-1}$, which also converges by the Alternating Series Test. Thus the interval of convergence is $[-1,1]$.

(3) Let $\displaystyle a_n=\frac{1}{n!}$. Then $$\lim_{n\to\infty}\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n\to\infty}\frac{|x|}{n+1}=0 \lt 1$$ for every $x$, so the series converges absolutely for every real $x$. The radius of convergence is $R=\infty$ and the interval of convergence is $(-\infty,\infty)$.

(4) Let $\displaystyle a_n=n!$. For $x\neq 0$, $$\lim_{n\to\infty}\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \lim_{n\to\infty}(n+1)|x| = \infty,$$ so the series diverges for $x\neq0$. At $x=0$ the series converges to $0$. Hence the radius of convergence is $R=0$, and the series converges only at $x=0$.

(5) Let $\displaystyle a_n=\frac{(-1)^n}{\sqrt{n+3}}$. Then $$\lim_{n\to\infty}\left|\frac{a_{n+1}x^{n+2}}{a_nx^{n+1}}\right| = \lim_{n\to\infty}\sqrt{\frac{n+3}{n+4}}|x| = |x|,$$ so the series converges absolutely for $|x| \lt 1$ and diverges for $|x| \gt 1$. The radius of convergence is $R=1$.

At $x=1$, the series is $\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{\sqrt{n+3}}$, which converges by the Alternating Series Test since $1/\sqrt{n+3}$ decreases monotonically to $0$. At $x=-1$, the series is $$\sum_{n=0}^{\infty}\frac{(-1)^n(-1)^{n+1}}{\sqrt{n+3}}=\sum_{n=0}^{\infty}\left(-\frac{1}{\sqrt{n+3}}\right),$$ which is $-1$ times a $p$-series with $p=1/2 \lt 1$ and hence diverges. Thus the interval of convergence is $(-1,1]$.

(6) Let $\displaystyle a_n=\frac{1}{n(\ln n)^2}$. Then $$\lim_{n\to\infty}\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right|=\lim_{n\to\infty}\frac{n(\ln n)^2}{(n+1)(\ln(n+1))^2}|x|=|x|,$$ so the series converges absolutely for $|x| \lt 1$ and diverges for $|x| \gt 1$. The radius of convergence is $R=1$.

At $x=1$, the series is $\displaystyle\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^2}$; by the Cauchy condensation test it has the same convergence behavior as $\displaystyle\sum_{n=2}^{\infty}2^n\cdot\frac{1}{2^n(\ln 2^n)^2}=\sum_{n=2}^{\infty}\frac{1}{n^2(\ln2)^2}$, a constant multiple of a $p$-series with $p=2 \gt 1$, so it converges. At $x=-1$, the series is $\displaystyle\sum_{n=2}^{\infty}\frac{(-1)^n}{n(\ln n)^2}$, which converges absolutely since its series of absolute values converges as shown above. Thus the interval of convergence is $[-1,1]$.

Solution.

(1) Let $\displaystyle a_n=\left(\frac{n}{n+1}\right)^{n^2}$. By the Root Test, $$\sqrt[n]{|a_n|} = \left(\frac{n}{n+1}\right)^n = \left(1-\frac{1}{n+1}\right)^n.$$ Since $\displaystyle\left(1-\frac{1}{n+1}\right)^{n+1}\to e^{-1}$, $$\left(1-\frac{1}{n+1}\right)^n = \left(1-\frac{1}{n+1}\right)^{n+1}\cdot\left(1-\frac{1}{n+1}\right)^{-1}\to e^{-1}\cdot1=e^{-1}.$$ Hence $$\lim_{n\to\infty}\sqrt[n]{|a_nx^n|} = e^{-1}|x|,$$ and the series converges when this is less than $1$, i.e. $|x| \lt e$. The radius of convergence is $R=e$.

(2) Let $\displaystyle c_n = \frac{2\cdot4\cdot6\cdots(2n)}{2\cdot5\cdot8\cdots(3n-1)}$ and $a_n=c_n^2$. By the Ratio Test, $$\frac{a_{n+1}}{a_n} = \left(\frac{c_{n+1}}{c_n}\right)^2 = \left(\frac{2(n+1)}{3(n+1)-1}\right)^2 = \left(\frac{2n+2}{3n+2}\right)^2 \to \left(\frac23\right)^2=\frac49,$$ so $$\lim_{n\to\infty}\left|\frac{a_{n+1}x^{n+1}}{a_nx^n}\right| = \frac49|x|.$$ The series converges when this is less than $1$, i.e. $|x| \lt \dfrac94$. The radius of convergence is $\displaystyle R=\frac94$.

Solution. The first series is $\displaystyle\sum_{n=0}^{\infty}x^n = 1+x+x^2+x^3+\cdots$, with every coefficient $a_n=1$. The second series is $\displaystyle\sum_{n=0}^{\infty}(-1)^n\frac{x^{n+1}}{n+1}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$; substituting $m=n+1$, the coefficient of $x^m$ is $\displaystyle b_m=\frac{(-1)^{m-1}}{m}$ for $m\ge1$, and $b_0=0$.

The product of the two series is given by the Cauchy product $$\left(\sum_{n=0}^{\infty}a_nx^n\right)\left(\sum_{m=0}^{\infty}b_mx^m\right) = \sum_{k=0}^{\infty}c_kx^k, \qquad c_k=\sum_{j=0}^{k}a_jb_{k-j},$$ and since $a_j=1$, $$c_k = \sum_{j=0}^{k}b_{k-j} = \sum_{m=0}^{k}b_m = \sum_{m=1}^{k}\frac{(-1)^{m-1}}{m}.$$

Hence $$c_0=0,\qquad c_1=1,\qquad c_2=1-\frac12=\frac12,\qquad c_3=1-\frac12+\frac13=\frac56,$$ so the product, up to the cubic term, is $$x+\frac12x^2+\frac56x^3+\cdots.$$

Solution. For $|x| \lt 1$, $\displaystyle\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$, and this series may be differentiated or integrated term by term within its interval of convergence.

(1) Differentiating both sides term by term with respect to $x$, $$\sum_{n=1}^{\infty}nx^{n-1} = \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2}.$$

(2) Differentiating both sides of (1) term by term once more, $$\sum_{n=2}^{\infty}n(n-1)x^{n-2} = \frac{d}{dx}\left(\frac{1}{(1-x)^2}\right) = \frac{2}{(1-x)^3}.$$

(3) Substituting $-x^2$ for $x$ in $\displaystyle\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ gives, for $|x| \lt 1$, $$\sum_{n=0}^{\infty}(-1)^nx^{2n} = \frac{1}{1+x^2}.$$ Integrating this series term by term from $0$ to $x$, $$\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{2n+1} = \int_0^x\frac{1}{1+t^2}\,dt = \tan^{-1}x.$$

(4) Substituting $-x$ for $x$ in $\displaystyle\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ gives, for $|x| \lt 1$, $$\sum_{n=0}^{\infty}(-1)^nx^n = \frac{1}{1+x}.$$ Integrating this series term by term from $0$ to $x$, $$\sum_{n=0}^{\infty}\frac{(-1)^nx^{n+1}}{n+1} = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^n}{n} = \int_0^x\frac{1}{1+t}\,dt = \ln(1+x).$$