Exercises

Problems to check your understanding of this chapter. The solutions are folded away — try them yourself first.

Easy

1Easy

Find the radius of convergence of the power series $\sum_{n=0}^{\infty}\frac{n^2+1}{3^n}\,x^n$ .

\left|\frac{c_{n+1}x^{n+1}}{c_n x^n}\right| = \frac{(n+1)^2+1}{n^2+1}\cdot\frac{|x|}{3} \to \frac{|x|}{3}

Power Series and Radius of Convergence

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With $c_n = \frac{n^2+1}{3^n} > 0$ , the ratio of successive absolute terms is $\left|\frac{c_{n+1}x^{n+1}}{c_n x^n}\right| = \frac{(n+1)^2+1}{n^2+1}\cdot\frac{|x|}{3}$ .
Since $\frac{(n+1)^2+1}{n^2+1} = \frac{n^2+2n+2}{n^2+1} \to 1$ , the ratio tends to $\frac{|x|}{3}$ .
By the ratio test the series converges (absolutely) for $|x| < 3$ , and for $|x| > 3$ the general term $\left|c_n x^n\right| = (n^2+1)\left(\frac{|x|}{3}\right)^n \to \infty$ , so it diverges. Hence the radius of convergence is $R = 3$ .

Medium

2Medium

Prove that $\sum_{n=1}^{\infty} n(-1)^n$ diverges in the ordinary sense but is Abel summable, with Abel sum $-\frac{1}{4}$ .

\sum_{n\ge 1} n(-1)^n x^n = -\frac{x}{(1+x)^2} \xrightarrow[x\to 1^{-}]{} -\frac{1}{4}

Abel's Theorem

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Divergence first: the general term $n(-1)^n$ does not tend to $0$ (since $|n(-1)^n| = n \to \infty$ ), so the series diverges by the $n$ -th term test.
For $|z| < 1$ , differentiating $\sum_{n=0}^{\infty} z^n = \frac{1}{1-z}$ termwise (a power series may be differentiated termwise inside its disc) gives $\sum_{n=1}^{\infty} n z^{n-1} = \frac{1}{(1-z)^2}$ ; multiplying by $z$ yields $\sum_{n=1}^{\infty} n z^n = \frac{z}{(1-z)^2}$ .
Substituting $z = -x$ ( $0 \le x < 1$ ) gives $n(-x)^n = n(-1)^n x^n$ , so $\sum_{n=1}^{\infty} n(-1)^n x^n = \frac{-x}{(1+x)^2}$ . As $x \to 1^{-}$ this tends to $-\frac{1}{4}$ , the Abel sum. A divergent series can still be Abel summable.

3Medium

Prove rigorously that the rearrangement of the alternating harmonic series taking "two positive terms, then one negative term" in order, $1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots$ , converges to $\frac{3}{2}\ln 2$ . (You may use $H_m = \ln m+\gamma+o(1)$ as known.)

T_{3N} = H_{4N}-\tfrac12 H_{2N}-\tfrac12 H_N \to \tfrac32\ln 2

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This is a genuine rearrangement: the odd reciprocals $\frac{1}{2k-1}$ and the even reciprocals $\frac{1}{2k}$ each appear exactly once, in order. Let $T_{3N}$ be the sum of the first $3N$ terms; the positive terms used are the first $2N$ odd reciprocals and the negative terms the first $N$ even reciprocals, so $T_{3N} = \sum_{k=1}^{2N}\frac{1}{2k-1}-\sum_{k=1}^{N}\frac{1}{2k}$ .
Since $\sum_{k=1}^{2N}\frac{1}{2k-1} = H_{4N}-\sum_{k=1}^{2N}\frac{1}{2k} = H_{4N}-\frac{1}{2}H_{2N}$ , we get $T_{3N} = H_{4N}-\frac{1}{2}H_{2N}-\frac{1}{2}H_N$ .
Substituting $H_m = \ln m+\gamma+\varepsilon_m$ ( $\varepsilon_m \to 0$ ) gives $T_{3N} = \ln(4N)-\frac{1}{2}\ln(2N)-\frac{1}{2}\ln N+\gamma\left(1-\frac{1}{2}-\frac{1}{2}\right)+o(1)$ . The $\gamma$ terms cancel and the logarithms give $\ln 4-\frac{1}{2}\ln 2 = \frac{3}{2}\ln 2$ , so $T_{3N} \to \frac{3}{2}\ln 2$ .
The intermediate partial sums follow: $T_{3N+1} = T_{3N}+\frac{1}{4N+1}$ and $T_{3N+2} = T_{3N+1}+\frac{1}{4N+3}$ , and the added terms tend to $0$ , so the whole sequence $T_m$ converges to $\frac{3}{2}\ln 2$ . A conditionally convergent series can change its sum under rearrangement.