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Worked solution

Check powers before taking an infinite GP sum

For the geometric partial sum Sn=a(1−rn)1−rS_n=\frac{a(1-r^n)}{1-r}, consider what happens to rnr^n as n→∞n\to\infty and whether there is a finite sum to infinity. Assume a≠0a\ne0; treat r=1r=1 separately from this formula.

CJC · Sequences and Series · What happens to the ratio power?

Answer

A nonzero geometric series converges exactly when ∣r∣<1|r|<1, with S∞=a1−rS_\infty=\frac{a}{1-r}.

Compare your reasoning

First inspect ratios of modulus below one

∣r∣<1⟹rn→0,Sn→a1−r|r|<1\quad\Longrightarrow\quad r^n\to0,\qquad S_n\to\frac{a}{1-r}

The source identifies −1<r<1 and takes the power to zero. This includes decaying alternating powers when r is negative.

Separate the two ways a power can grow

r>1: rn→+∞;r<−1: ∣rn∣→+∞ with alternating signr>1:\ r^n\to+\infty;\qquad r<-1:\ |r^n|\to+\infty\ \text{with alternating sign}

For r>1, r^n grows positively; the sign of the divergent sum follows a. For r<−1 the powers alternate with increasing magnitude, so neither r^n nor the nonzero partial sums tend to positive infinity. Both cases diverge.

Check the boundary ratios

r=1: Sn=na;r=−1: S2k=0, S2k−1=ar=1:\ S_n=na;\qquad r=-1:\ S_{2k}=0,\ S_{2k-1}=a

These authored boundary cases complete the criterion. At r=1 use the constant sum; at r=−1 the partial sums alternate between a and zero.

How did this question go?

Solution notes

The source prints r^n→∞ and S_n→∞ for |r|>1. That direction needs correction for negative ratios and for a negative first term. Divergence is still the correct conclusion for every nonzero a with |r|>1. The all-zero extension converges for any ratio.