[EXAMPLES] Power series

Consider

\sum_{k=0}^{\infty}\frac{(x-3)^k}{3^k+2}.

Writing its term as $a_k$ , the ratio test gives

\left|\frac{a_{k+1}}{a_k}\right| =|x-3|\frac{3^k+2}{3^{k+1}+2} \longrightarrow\frac{|x-3|}{3}.

Therefore the series converges when $|x-3|<3$ , giving the possible interval $(0,6)$ .

At $x=0$ ,

a_k=\frac{(-3)^k}{3^k+2}, \qquad |a_k|\longrightarrow1,

so the terms do not tend to zero. At $x=6$ ,

a_k=\frac{3^k}{3^k+2}\longrightarrow1,

so the terms again fail the $n$ -th term test. The interval of convergence is

{(0,6)}.

[exam]

Consider

\sum_{k=1}^{\infty}\frac{4^k}{k}(x-2)^k.

The ratio test gives

\left|\frac{a_{k+1}}{a_k}\right| =4|x-2|\frac{k}{k+1} \longrightarrow4|x-2|.

Thus $R=1/4$ , with possible interval $(7/4,9/4)$ .

At $x=7/4$ ,

\frac{4^k}{k}\left(-\frac14\right)^k =\frac{(-1)^k}{k},

which converges by the alternating-series test. At $x=9/4$ ,

\frac{4^k}{k}\left(\frac14\right)^k =\frac1k,

which diverges. Hence

{\left[\frac74,\frac94\right)}.

[exam]

Termwise differentiation of

\sin x=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)!}

gives

\cos x =\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k}}{(2k)!}.

[routine]

Start with the geometric series:

\frac1{1-x}=\sum_{k=0}^{\infty}x^k, \qquad |x|<1.

Differentiate and multiply by $x$ :

\frac{x}{(1-x)^2}=\sum_{k=0}^{\infty}kx^k.

Differentiate once more and multiply by $x$ :

\frac{x(1+x)}{(1-x)^3} =\sum_{k=0}^{\infty}k^2x^k, \qquad |x|<1.

Therefore

{ \sum_{k=0}^{\infty}k^2x^k =\frac{x(1+x)}{(1-x)^3}, \qquad |x|<1. }

[exam]

Examples from Power series