Exercises — Arithmetic-geometric progression — finding sum

What we do: split each term into (AP number) $\times$ (GP number).

• AP part: the numbers multiplying the powers are $5,8,11,14,\dots$ — they go up by $3$. So $a=5$, $d=3$.
• GP part: the powers are $1,\frac{1}{3},\frac{1}{9},\frac{1}{27},\cdots ={\left(\frac{1}{3}\right)}^0,{\left(\frac{1}{3}\right)}^1,\dots$ So $r=\frac{1}{3}$.

General term (AP value $\times$ GP value): $T_n=(5+(n-1)\cdot 3){\left(\frac{1}{3}\right)}^{n-1}=(3n+2){\left(\frac{1}{3}\right)}^{n-1}.$ Answer: $a=5,d=3,r=\frac{1}{3},T_n=(3n+2)/3^{n-1}$.

What we do: try to write $T_n=(a+(n-1)d)r^{n-1}$. Take $r=2$: then $r^{n-1}=1,2,4,8,\dots$. To match $2,4,8,16$ we need the AP part to be constant $=2$, i.e. $a=2$ and $d=0$. So yes — it is an AGP with $a=2,d=0,r=2$. When $d=0$ the arithmetic part never changes, so the AGP collapses into a pure GP (see Sum to infinity of GP). AGP is the general family; GP is the $d=0$ corner of it.

Extract: coefficients $1,3,5,7\Rightarrow a=1,d=2$; powers give $r=x$. Since $|x|<1$, the infinite formula is legal. $S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}=\frac{1}{1-x}+\frac{2x}{(1-x{)}^2}.$ Combine over $(1-x{)}^2$: $=\frac{(1-x)+2x}{(1-x{)}^2}=\frac{1+x}{(1-x{)}^2}.$ Answer: $\frac{1+x}{(1-x{)}^2}$.

Extract: $a=1,d=1$; the powers $3^{-(n-1)}$ mean $r=\frac{1}{3}$. Here $|r|=\frac{1}{3}<1$, so the infinite sum exists. $S=\frac{1}{1-\frac{1}{3}}+\frac{1\cdot \frac{1}{3}}{{\left(1-\frac{1}{3}\right)}^2}=\frac{1}{\frac{2}{3}}+\frac{\frac{1}{3}}{\frac{4}{9}}.$ First piece $=\frac{3}{2}$. Second piece $=\frac{1}{3}\cdot \frac{9}{4}=\frac{3}{4}$. $S=\frac{3}{2}+\frac{3}{4}=\frac{9}{4}.$ Answer: $\frac{9}{4}=2.25$. (Sanity: partial sums $1,1.667,1.999,2.147,\cdots \to 2.25$.)

Direct add: the terms are $1\cdot 1=1,2\cdot 2=4,3\cdot 4=12,4\cdot 8=32$, so $1+4+12+32=49$. Formula (note $r=2>1$ — only the finite formula is allowed here). With $a=1,d=1,r=2,n=4$: $S_4=\frac{1}{1-2}+\frac{1\cdot 2(1-2^3)}{(1-2{)}^2}-\frac{(1+3\cdot 1)2^4}{1-2}.$

• First term: $\frac{1}{-1}=-1$.
• Second: $\frac{2(1-8)}{(-1{)}^2}=\frac{2\cdot (-7)}{1}=-14$.
• Third: $-\frac{4\cdot 16}{-1}=+64$. $S_4=-1-14+64=49.✓$

The disguise: the power is $5^{-n}$, not $5^{-(n-1)}$. Our formula assumes $r^{n-1}$. Factor one power out to fix the offset: $S={\sum }_{n=1}^{\infty }(2n-1){\left(\frac{1}{5}\right)}^n=\frac{1}{5}{\sum }_{n=1}^{\infty }(2n-1){\left(\frac{1}{5}\right)}^{n-1}.$ Now it is standard: inside, $a=1,d=2,r=\frac{1}{5}$. $\sum (2n-1)r^{n-1}=\frac{1}{1-\frac{1}{5}}+\frac{2\cdot \frac{1}{5}}{{\left(1-\frac{1}{5}\right)}^2}=\frac{1}{\frac{4}{5}}+\frac{\frac{2}{5}}{\frac{16}{25}}=\frac{5}{4}+\frac{2}{5}\cdot \frac{25}{16}=\frac{5}{4}+\frac{5}{8}=\frac{15}{8}.$ Multiply back by $\frac{1}{5}$: $S=\frac{1}{5}\cdot \frac{15}{8}=\frac{3}{8}.$ Answer: $\frac{3}{8}$.

Set up: $\frac{1}{(1-r{)}^2}=4\Rightarrow (1-r{)}^2=\frac{1}{4}\Rightarrow 1-r=\pm \frac{1}{2}.$

• $1-r=+\frac{1}{2}\Rightarrow r=\frac{1}{2}$. This satisfies $|r|<1$. ✓
• $1-r=-\frac{1}{2}\Rightarrow r=\frac{3}{2}$. Here $|r|=\frac{3}{2}>1$, so the infinite series diverges — the algebraic root is spurious. Answer: only $r=\frac{1}{2}$ is valid. (Check: $\frac{1}{(1-\frac{1}{2}{)}^2}=\frac{1}{1/4}=4$. ✓)

Why the hint? Our AGP formula handles $\sum n{r}^{n-1}$ (linear coefficient), not $\sum n^2{r}^{n-1}$. But $n^2=n(n-1)+n$ turns the quadratic into pieces we can sum. With $r=\frac{1}{2}$: $S=\underset{\text{Piece A}}{\underbrace{\sum n(n-1)r^{n-1}}}+\underset{\text{Piece B}}{\underbrace{\sum n{r}^{n-1}}}.$

Piece B first (re-derived here, not just cited). Let $B={\sum }_{n\ge 1}n{r}^{n-1}$. Use the same multiply-and-shift trick as an AGP with $a=1,d=1$: $B=1+2r+3r^2+4r^3+\cdots ,rB=r+2r^2+3r^3+\cdots$ Subtract: $B-rB=1+r+r^2+r^3+\cdots =\frac{1}{1-r}$ (a pure GP, $|r|<1$). So $(1-r)B=\frac{1}{1-r}$, giving $B=\frac{1}{(1-r{)}^2}.$ At $r=\frac{1}{2}$: $B=\frac{1}{(1/2{)}^2}=4.$

Piece A (justify the identity on-page). Let $A_0={\sum }_{n\ge 1}n(n-1)r^{n-2}$ (I pulled out $r^{n-2}$ so the counting is clean). Watch WHY this equals $\frac{2}{(1-r{)}^3}$: differentiate the geometric series twice. Start from $\frac{1}{1-r}={\sum }_{n\ge 0}{r}^n=1+r+r^2+r^3+\cdots$ Differentiate once w.r.t. $r$ (each $r^n\to n{r}^{n-1}$): $\frac{1}{(1-r{)}^2}={\sum }_{n\ge 1}n{r}^{n-1}.$ Differentiate again (each $n{r}^{n-1}\to n(n-1)r^{n-2}$): $\frac{2}{(1-r{)}^3}={\sum }_{n\ge 2}n(n-1)r^{n-2}.$ Why differentiation is the right tool: each derivative in $r$ pulls the exponent down as a multiplying factor, exactly manufacturing the coefficients $n$ then $n(n-1)$ we need — it is the machine that turns a plain GP into these weighted sums (see Power series and generating functions). So $A_0=\frac{2}{(1-r{)}^3}$. Now Piece A wanted $r^{n-1}$, one power higher than $A_0$, so Piece A $=r\cdot A_0$: $\text{Piece A}=r\cdot \frac{2}{(1-r{)}^3}=\frac{1}{2}\cdot \frac{2}{(1/2{)}^3}=\frac{1}{2}\cdot \frac{2}{1/8}=\frac{1}{2}\cdot 16=8.$

Combine: $S=\text{Piece A}+\text{Piece B}=8+4=12.$ Answer: $S=12$. (Sanity: partial sums $1,3,5.25,7.25,8.8,9.9,\cdots \to 12$.)

Why this is different: with $r=-\frac{1}{2}$ the powers flip sign every term, so the series alternates $+,-,+,-$. The formula still applies because the only requirement is $|r|<1$, and $|-\frac{1}{2}|=\frac{1}{2}<1$. ✓ Extract: $a=1,d=1,r=-\frac{1}{2}$. Note $1-r=1-(-\frac{1}{2})=\frac{3}{2}$. $S=\frac{1}{1-r}+\frac{dr}{(1-r{)}^2}=\frac{1}{\frac{3}{2}}+\frac{1\cdot (-\frac{1}{2})}{{\left(\frac{3}{2}\right)}^2}=\frac{2}{3}+\frac{-\frac{1}{2}}{\frac{9}{4}}.$ Second piece $=-\frac{1}{2}\cdot \frac{4}{9}=-\frac{2}{9}$. $S=\frac{2}{3}-\frac{2}{9}=\frac{6}{9}-\frac{2}{9}=\frac{4}{9}.$ Answer: $S=\frac{4}{9}$. (Sanity: partial sums $1,0,0.75,0.25,0.5625,\dots$ oscillate toward $0.444\dots =\frac{4}{9}$ — the tug-of-war between $+$ and $-$ terms settles down because each is smaller than the last.)

Plug in ($1-r=\frac{1}{2}$): $S_5=\frac{2}{\frac{1}{2}}+\frac{3\cdot \frac{1}{2}\left(1-{\left(\frac{1}{2}\right)}^4\right)}{{\left(\frac{1}{2}\right)}^2}-\frac{(2+4\cdot 3){\left(\frac{1}{2}\right)}^5}{\frac{1}{2}}.$

• Term 1: $2/\frac{1}{2}=4$.
• Term 2: numerator $=\frac{3}{2}\left(1-\frac{1}{16}\right)=\frac{3}{2}\cdot \frac{15}{16}=\frac{45}{32}$; divide by $\frac{1}{4}$: $\frac{45}{32}\cdot 4=\frac{45}{8}$.
• Term 3: $(2+12)=14$; $14\cdot \frac{1}{32}=\frac{14}{32}=\frac{7}{16}$; divide by $\frac{1}{2}$ (with the minus sign): $-\frac{7}{16}\cdot 2=-\frac{7}{8}$. $S_5=4+\frac{45}{8}-\frac{7}{8}=4+\frac{38}{8}=4+\frac{19}{4}=\frac{35}{4}.$ Answer: $S_5=\frac{35}{4}=8.75$. (Direct check: $2+2.5+2+1.375+0.875=8.75$. ✓)

Write $S=3+5\cdot \frac{1}{4}+7\cdot \frac{1}{16}+9\cdot \frac{1}{64}+\cdots$ (coefficients $3,5,7,9,\dots$, step $d=2$). Multiply by $r=\frac{1}{4}$ and shift: $\frac{1}{4}S=3\cdot \frac{1}{4}+5\cdot \frac{1}{16}+7\cdot \frac{1}{64}+\cdots$ Subtract: for each aligned power the coefficient drops by $d=2$, and the very first $3$ has nothing above it: $S-\frac{1}{4}S=3+(2\cdot \frac{1}{4}+2\cdot \frac{1}{16}+2\cdot \frac{1}{64}+\cdots ).$ The bracket is a pure GP with first term $2\cdot \frac{1}{4}=\frac{1}{2}$, ratio $\frac{1}{4}$: its sum is $\frac{1/2}{1-1/4}=\frac{1/2}{3/4}=\frac{2}{3}$. $\frac{3}{4}S=3+\frac{2}{3}=\frac{11}{3}\Rightarrow S=\frac{4}{3}\cdot \frac{11}{3}=\frac{44}{9}.$ Formula check: $\frac{3}{3/4}+\frac{2\cdot \frac{1}{4}}{(3/4{)}^2}=4+\frac{1/2}{9/16}=4+\frac{8}{9}=\frac{44}{9}.✓$ Answer: $S_{\infty }=\frac{44}{9}$.

Why the formula dies: every denominator contains $(1-r)$; at $r=1$ that is $0$ — division by zero. The multiply-and-shift trick also fails because multiplying by $r=1$ changes nothing, so there is nothing to cancel. What actually happens: with $r=1$ each GP factor is $1$, so $T_n=a+(n-1)d$ — the series is just a pure AP (see Arithmetic Progression). Use the AP sum: $S_6=\frac{n}{2}(2a+(n-1)d)=\frac{6}{2}(2\cdot 2+5\cdot 3)=3\cdot (4+15)=3\cdot 19=57.$ Answer: $S_6=57$. The AGP method degenerates to the AP formula when $r=1$.

Why this matters: $S_{\infty }$ is obtained by dropping this "leftover" third term as $n\to \infty$. We watch it die. Here $\frac{1}{1-r}=\frac{3}{2}$, so the third term is $-\frac{3}{2}n{\left(\frac{1}{3}\right)}^n$.

• $n=1$: $-\frac{3}{2}\cdot 1\cdot \frac{1}{3}=-\frac{1}{2}=-0.5$.
• $n=3$: $-\frac{3}{2}\cdot 3\cdot \frac{1}{27}=-\frac{3}{2}\cdot \frac{3}{27}=-\frac{1}{6}\approx -0.1667$.
• $n=10$: $-\frac{3}{2}\cdot 10\cdot \frac{1}{59049}\approx -2.54\times 10^{-4}$. Trend: $-0.5,-0.167,-0.00025,\cdots \to 0$. The exponential $r^n$ crushes the linear $n$ (this is exactly Limits: exponential vs polynomial growth). Hence the finite sum slides into $S_{\infty }=\frac{1}{1-\frac{1}{3}}+\frac{\frac{1}{3}}{(1-\frac{1}{3}{)}^2}=\frac{3}{2}+\frac{3}{4}=\frac{9}{4}$. Conclusion: the third term decays to $0$ as $n\to \infty$ (values $-0.5,-\frac{1}{6},\approx -0.000254$), so $S_n\to S_{\infty }=\frac{9}{4}$. This is precisely why the infinite formula is just the finite one with its last two $r^n$-carrying pieces deleted. Answer: third term $=-0.5,-\frac{1}{6},\approx -0.000254$; limit $0$; $S_{\infty }=\frac{9}{4}$.