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Worked solution

Classify infinite series from their partial sums

Classify the series represented by these partial sums as convergent or divergent, giving the finite limit when one exists. (a) ∑j=1n2j=2(2n−1)\sum_{j=1}^{n}2^j=2(2^n-1). (b) Given ∑j=1n1j(j+2)=34−12(n+1)−12(n+2)\sum_{j=1}^{n}\frac1{j(j+2)}=\frac34-\frac1{2(n+1)}-\frac1{2(n+2)}. (c) An AP with Sn=n2[2a+(n−1)d]S_n=\frac n2[2a+(n-1)d], for d>0d>0 and d<0d<0. (d) A GP with Sn=a(rn−1)r−1S_n=\frac{a(r^n-1)}{r-1}, r>1r>1, a≠0a\ne0. (e) A GP with Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}, ∣r∣<1|r|<1.

CJC · Sequences and Series · Convergence comparison table

(a) Increasing powers of two

Answer

Divergent; the partial sums tend to positive infinity.

Inspect the growing power

2n→+∞⟹2(2n−1)→+∞2^n\to+\infty\quad\Longrightarrow\quad2(2^n-1)\to+\infty

The table uses 2^n tending to infinity.

How did this part go?

(b) Reciprocal products

Answer

Convergent with sum 34\frac34.

Let both reciprocal remainders tend to zero

lim⁡n→∞[34−12(n+1)−12(n+2)]=34\begin{aligned}&\lim_{n\to\infty}\left[\frac34-\frac1{2(n+1)}-\frac1{2(n+2)}\right]\\ & =\frac34\end{aligned}

The Teacher fills in the classification and the finite sum.

How did this part go?

(c) AP with nonzero difference

Answer

Divergent: positive infinity for d>0d>0, negative infinity for d<0d<0.

Expand the partial sum

Sn=d2n2+(a−d2)nS_n=\frac d2n^2+\left(a-\frac d2\right)n

The quadratic term dominates the linear term for fixed a and nonzero d.

Keep the sign of the leading coefficient

d>0⇒Sn→+∞,d<0⇒Sn→−∞d>0\Rightarrow S_n\to+\infty,\qquad d<0\Rightarrow S_n\to-\infty

These are the two cases explicitly printed in the table.

How did this part go?

(d) GP with ratio greater than one

Answer

Divergent: positive infinity for a>0a>0, negative infinity for a<0a<0.

Track both the power and the first-term sign

rn→+∞,ar−1>0 if a>0,ar−1<0 if a<0r^n\to+\infty,\qquad\frac{a}{r-1}>0\ \text{if }a>0,\qquad\frac{a}{r-1}<0\ \text{if }a<0

The source prints positive infinity without specifying a positive first term. The conclusion of divergence is unchanged for every nonzero a, but its direction depends on a.

How did this part go?

(e) GP with ratio of modulus below one

Answer

Convergent with sum a1−r\frac{a}{1-r}.

Use decay of the power

∣r∣<1⟹rn→0⟹lim⁡n→∞a(1−rn)1−r=a1−r|r|<1\quad\Longrightarrow\quad r^n\to0\quad\Longrightarrow\quad\lim_{n\to\infty}\frac{a(1-r^n)}{1-r}=\frac{a}{1-r}

A negative ratio of modulus below one alternates in sign but still has r^n tending to zero.

How did this part go?

Solution notes

The source table supplies the partial-sum working and leaves the Student classification cells blank. The Teacher fills those cells. The GP with r>1 is qualified here for the sign of a; a nonzero GP is assumed. The all-zero extension is handled separately in the constant-case activity.