Exercises — Sum of infinite GP — when it converges, proof

We test whether the ratio $\frac{\text{next}}{\text{current}}$ is the same every step. That is what makes a list geometric.

• (a) $10/5=2$, $20/10=2$, $40/20=2$. Same ratio → GP with $a=5,r=2$. Since $|r|=2\ge 1$, diverges.
• (b) $3/9=\frac{1}{3}$, $1/3=\frac{1}{3}$, $(1/3)/1=\frac{1}{3}$. GP with $a=9,r=\frac{1}{3}$. $|r|<1$ → converges.
• (c) $4/2=2$ but $6/4=1.5$. Ratio changes → not a GP (it's arithmetic: adding $2$).
• (d) $-4/8=-\frac{1}{2}$, $2/-4=-\frac{1}{2}$, $-1/2=-\frac{1}{2}$. GP with $a=8,r=-\frac{1}{2}$. $|r|=\frac{1}{2}<1$ → converges.

First term you see: $a=12$. Ratio: $\frac{4}{12}=\frac{1}{3}$; check next $\frac{4/3}{4}=\frac{1}{3}$. So $r=\frac{1}{3}$. Since $|r|=\frac{1}{3}<1$, it converges.

$a=6$, $r=\frac{2}{6}=\frac{1}{3}$. Check: $|r|=\frac{1}{3}<1$ → converges, so the formula is legal. $S_{\infty }=\frac{a}{1-r}=\frac{6}{1-\frac{1}{3}}=\frac{6}{\frac{2}{3}}=6\cdot \frac{3}{2}=9.$

Signs alternate, so $r<0$. $a=3$, $r=\frac{-1}{3}=-\frac{1}{3}$. $|r|=\frac{1}{3}<1$ → converges. $S_{\infty }=\frac{3}{1-(-\frac{1}{3})}=\frac{3}{\frac{4}{3}}=3\cdot \frac{3}{4}=\frac{9}{4}.$

We know $S_{\infty }=\frac{a}{1-r}$. Rearrange for $r$: multiply both sides by $(1-r)$, $8(1-r)=10⟹1-r=\frac{10}{8}=\frac{5}{4}⟹r=1-\frac{5}{4}=-\frac{1}{4}.$ Check: $|r|=\frac{1}{4}<1$ ✓ so this GP genuinely converges, answer valid.

This is a GP with $a=1$ and $r=x$. Convergence needs $|r|<1$, i.e. $==-1<x<1==$. On that interval, $S_{\infty }=\frac{1}{1-x}.$ At $x=1$: sum is $1+1+1+\cdots \to \infty$. At $x=-1$: partial sums are $1,0,1,0,\dots$ — they oscillate and never settle. Both endpoints are excluded.

Ratio: $\frac{2x}{1}=2x$, and $\frac{4x^2}{2x}=2x$ ✓. So $a=1$, $r=2x$. Converge condition: $|2x|<1⟹|x|<\frac{1}{2}$, i.e. $-\frac{1}{2}<x<\frac{1}{2}$. $S_{\infty }=\frac{1}{1-2x}.$

From $S_{\infty }=\frac{a}{1-r}$: $a=S_{\infty }(1-r)=15\left(1-\frac{2}{5}\right)=15\cdot \frac{3}{5}=9.$ Third term is $a{r}^2=9{\left(\frac{2}{5}\right)}^2=9\cdot \frac{4}{25}=\frac{36}{25}=\mathrm{1.44.}$

Break it into repeating blocks of two digits: $0.272727\dots =0.27+0.0027+0.000027+\cdots$ Each block is the previous one shifted two decimal places right → multiplied by $0.01$. So $a=0.27$, $r=0.01$, and $|r|<1$ so it converges. $S_{\infty }=\frac{0.27}{1-0.01}=\frac{0.27}{0.99}=\frac{27}{99}=\frac{3}{11}.$ See Recurring Decimals as Fractions for the general pattern.

Look at the figure. The outer black square has side $1$, so area $1$. The next (red) square joins midpoints; its side is the hypotenuse of a right triangle with legs $\frac{1}{2},\frac{1}{2}$, so side $=\sqrt{(\frac{1}{2}{)}^2+(\frac{1}{2}{)}^2}=\frac{1}{\sqrt{2}}$, and its area $={\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}$. Each step multiplies area by $\frac{1}{2}$, so areas form a GP: $1,\frac{1}{2},\frac{1}{4},\dots$ with $a=1,r=\frac{1}{2}$. Since $|r|<1$, $\text{Total area}=\frac{1}{1-\frac{1}{2}}=2.$

Payments: $100,90,81,\dots$ → GP with $a=100$, $r=0.9$. $|r|<1$ converges. $\text{Total}=\frac{100}{1-0.9}=\frac{100}{0.1}=1000.$

The squares $a^2,a^2{r}^2,a^2{r}^4,\dots$ form a new GP with first term $a^2$ and ratio $r^2$ (need $|r^2|<1$, automatic if $|r|<1$). So $\frac{a^2}{1-r^2}=12.$ Now use $a=6(1-r)$ from the first equation. Note $1-r^2=(1-r)(1+r)$, so $\frac{a^2}{(1-r)(1+r)}=\frac{[6(1-r){]}^2}{(1-r)(1+r)}=\frac{36(1-r{)}^2}{(1-r)(1+r)}=\frac{36(1-r)}{1+r}=12.$ Solve: $36(1-r)=12(1+r)⟹36-36r=12+12r⟹24=48r⟹r=\frac{1}{2}.$ Then $a=6(1-\frac{1}{2})=3.$ Check $|r|=\frac{1}{2}<1$ ✓.

(a) $S_3=\frac{4\left(1-(\frac{1}{3}{)}^3\right)}{1-\frac{1}{3}}=\frac{4\left(1-\frac{1}{27}\right)}{\frac{2}{3}}=\frac{4\cdot \frac{26}{27}}{\frac{2}{3}}=\frac{\frac{104}{27}}{\frac{2}{3}}=\frac{104}{27}\cdot \frac{3}{2}=\frac{312}{54}=\frac{52}{9}.$ (b) $S_{\infty }=\frac{4}{1-\frac{1}{3}}=\frac{4}{\frac{2}{3}}=6.$ (c) Gap $=S_{\infty }-S_3=6-\frac{52}{9}=\frac{54-52}{9}=\frac{2}{9}.$ Where does $\frac{2}{9}$ come from? From the algebra, $S_{\infty }-S_N=\frac{a}{1-r}-\frac{a(1-r^N)}{1-r}=\frac{a{r}^N}{1-r}$. With $N=3$: $\frac{4\cdot (\frac{1}{3}{)}^3}{\frac{2}{3}}=\frac{4/27}{2/3}=\frac{4}{27}\cdot \frac{3}{2}=\frac{2}{9}.$ ✓ This leftover $\frac{a{r}^N}{1-r}\to 0$ because $r^N\to 0$ — exactly the reason the infinite formula works. See Limits of Sequences.

First term $a=27$. Require convergence ($|r|<1$) and $S_{\infty }=2a$: $\frac{27}{1-r}=2\cdot 27=54⟹\frac{1}{1-r}=2⟹1-r=\frac{1}{2}⟹r=\frac{1}{2}.$ Check $|r|=\frac{1}{2}<1$ ✓. (In general, $S_{\infty }=ka$ with $k>0$ forces $1-r=\frac{1}{k}$, so $r=1-\frac{1}{k}$; convergence needs $k>\frac{1}{2}$.) Here the unique answer is $r=\frac{1}{2}$.

• L1 Recognition ::: same ratio every step = GP; converges iff $|r|<1$.
• L2 Application ::: check $|r|<1$, then $S_{\infty }=\frac{a}{1-r}$; rearrange to find $a$ or $r$.
• L3 Analysis ::: "for which $x$" gives a strict interval; endpoints $|r|=1$ diverge.
• L4 Synthesis ::: build the GP — decimals use $r=10^{-\text{block}}$; geometry/finance give $r$ from the picture/story.
• L5 Mastery ::: squaring a GP squares its ratio; the gap $S_{\infty }-S_N=\frac{a{r}^N}{1-r}\to 0$.