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Worked solution

Test convergence before evaluating geometric sums

For each geometric progression, determine the sum to infinity, if it exists: (i) 3,9,27,81,…3,9,27,81,\ldots; (ii) 0.9,0.92,0.93,0.94,…0.9,0.9^2,0.9^3,0.9^4,\ldots; (iii) 2,−43,89,−1627,…2,-\frac43,\frac89,-\frac{16}{27},\ldots.

CJC · Sequences and Series · Self-Practice 8

(i) Tripling terms

Answer

The sum to infinity does not exist.

Find and test the ratio

a=3,r=3,∣r∣>1a=3,\qquad r=3,\qquad |r|>1

The first term is 3 and the common ratio is 3. Its modulus exceeds one, so the nonzero series diverges.

How did this part go?

(ii) Powers of 0.9

Answer

99.

Keep the actual first term

a=0.9,r=0.9,∣r∣<1,S∞=0.91−0.9=9a=0.9,\quad r=0.9,\quad |r|<1,\qquad S_\infty=\frac{0.9}{1-0.9}=9

The printed progression starts at 0.9, not 1. The source gives 9; this working supplies the substitution.

How did this part go?

(iii) Alternating shrinking terms

Answer

65\frac65.

Retain the negative ratio

a=2,r=−23,∣r∣=23<1,S∞=21−(−2/3)=65a=2,\quad r=-\frac23,\quad |r|=\frac23<1,\qquad S_\infty=\frac2{1-(-2/3)}=\frac65

Divide the second term by the first: r=−2/3. Its modulus is below one, so the sum exists even though signs alternate.

How did this part go?

Common mistakes

Test |r|<1 before using a/(1−r). Use the signed ratio in the denominator; |r| is only for the convergence test.