Calculus III — Sequences & Series

Question 1

(a) [6] Write $S_n = a + ar + \dots + ar^{n-1}$. Multiply by $r$: $rS_n = ar + ar^2 + \dots + ar^{n}.$ Subtract: $S_n - rS_n = a - ar^n$, so $S_n = \frac{a(1-r^n)}{1-r}\quad (r\neq1).\;\text{[2]}$ Convergence: as $n\to\infty$, $r^n\to 0$ iff $|r|<1$; then $S_n\to \dfrac{a}{1-r}$. [2] If $|r|>1$, $|r^n|\to\infty$ (diverges); if $r=1$, $S_n=na\to\pm\infty$; if $r=-1$, $S_n$ oscillates $a,0,a,0,\dots$ — no limit. Hence convergence iff $|r|<1$ with sum $\frac{a}{1-r}$. [2]

(b) [4] $\sum_{n=1}^\infty\frac{3^n+2^n}{6^n}=\sum\left(\tfrac{1}{2}\right)^n+\sum\left(\tfrac{1}{3}\right)^n.$ [2] Each geometric with first term = ratio: $\frac{1/2}{1-1/2}+\frac{1/3}{1-1/3}=1+\frac12=\frac32.\;\text{[2]}$

Question 2

(a) [7] Since $f$ decreasing, on $[k,k+1]$: $f(k+1)\le f(x)\le f(k)$. Integrating over $[k,k+1]$: $a_{k+1}\le \int_k^{k+1} f\,dx \le a_k. \;\text{[3]}$ Summing $k=1$ to $n-1$: $\sum_{k=2}^{n} a_k \le \int_1^{n} f\,dx \le \sum_{k=1}^{n-1} a_k. \;\text{[2]}$ If $\int_1^\infty f$ converges, the partial sums $\sum_{k=2}^n a_k$ are bounded above and increasing ⇒ series converges. If $\int$ diverges, $\int_1^n f\to\infty$ forces $\sum_{k=1}^{n-1}a_k\to\infty$ ⇒ series diverges. Both directions established. [2]

(b) [5] Take $f(x)=x^{-p}$, positive/continuous/decreasing for $p>0$. $\int_1^\infty x^{-p}dx=\begin{cases}\dfrac{1}{p-1}& p>1\\ \infty & p\le1.\end{cases}\;\text{[2]}$ For $p\le 0$, $a_n\not\to0$ ⇒ diverges by divergence test. So converges iff $p>1$. [2] Boundary $p=1$: $\int_1^n \frac1x dx=\ln n\to\infty$, so harmonic series diverges. [1]

Question 3

(a) [6] $\dfrac{1}{n(n+2)}=\dfrac12\left(\dfrac1n-\dfrac1{n+2}\right).$ [2] $S_N=\frac12\sum_{n=1}^N\left(\frac1n-\frac1{n+2}\right)=\frac12\left(1+\frac12-\frac1{N+1}-\frac1{N+2}\right).\;\text{[3]}$ As $N\to\infty$: $S_N\to\frac12\cdot\frac32=\frac34.$ [1]

(b) [2] $a_n\to0$ only tells us the divergence test is inconclusive; convergence requires the partial sums to have a limit, which telescoping gives directly. [2]

Question 4

(a) [7] Given $a_n>0$, $a_n$ decreasing, $a_n\to0$, consider $S_N=\sum(-1)^{n+1}a_n$. Even partial sums: $S_{2m+2}-S_{2m}=a_{2m+1}-a_{2m+2}\ge0$ ⇒ $\{S_{2m}\}$ increasing. [2] Odd: $S_{2m+1}-S_{2m-1}=-a_{2m}+a_{2m+1}\le0$ ⇒ $\{S_{2m-1}\}$ decreasing. [1] Also $S_{2m}=S_{2m-1}-a_{2m}\le S_{2m-1}$, so even sums bounded above, odd below; both monotone-bounded ⇒ converge. Since $S_{2m+1}-S_{2m}=a_{2m+1}\to0$, common limit $S$. [2] Error: $S$ lies between consecutive partial sums, so $|S-S_N|\le|S_{N+1}-S_N|=a_{N+1}$. [2]

(b) [3] $\sum\frac{(-1)^{n+1}}{n}$: satisfies Leibniz ($1/n\downarrow0$) ⇒ converges. But $\sum\frac1n$ diverges (harmonic). Hence conditionally convergent. [3]

(c) [2] Need $a_{N+1}=\frac{1}{N+1}\le0.01$ ⇒ $N+1\ge100$ ⇒ $N\ge99$. So 99 terms. [2]

Question 5

(a) [6] Start with $\frac{1}{1+t}=\sum_{n=0}^\infty(-1)^n t^n$, $|t|<1$. [2] Integrate from $0$ to $x$ (valid termwise within radius): $\ln(1+x)=\int_0^x\frac{dt}{1+t}=\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots.\;\text{[3]}$ Interval of convergence: $(-1,1]$ (converges at $x=1$ by Leibniz, diverges at $x=-1$ → harmonic). [1]

(b) [4] $e^x-1-x=\frac{x^2}{2}+\frac{x^3}{6}+\cdots$; $1-\cos x=\frac{x^2}{2}-\frac{x^4}{24}+\cdots$. [2] $\frac{x^2/2+O(x^3)}{x^2/2+O(x^4)}\to 1.\;\text{[2]}$

Question 6

(a) [6] Ratio test: $\left|\frac{(x-2)^{n+1}/((n+1)3^{n+1})}{(x-2)^n/(n3^n)}\right|=\frac{n}{n+1}\cdot\frac{|x-2|}{3}\to\frac{|x-2|}{3}.$ [2] Converges for $|x-2|<3$ ⇒ $R=3$, interval kernel $(-1,5)$. [2] Endpoints: $x=5$: $\sum\frac1n$ diverges. $x=-1$: $\sum\frac{(-1)^n}{n}$ converges (Leibniz). Interval $[-1,5)$. [2]

(b) [2] Ratio test. Limitation: inconclusive when the limit $=1$ (e.g. at endpoints / p-series), requiring separate analysis. [2]

[
{"claim":"Q1b sum = 3/2","code":"n=symbols('n'); s=summation(Rational(1,2)**n,(n,1,oo))+summation(Rational(1,3)**n,(n,1,oo)); result = (s==Rational(3,2))"},
{"claim":"Q3 telescoping series sum = 3/4","code":"n=symbols('n'); s=summation(1/(n*(n+2)),(n,1,oo)); result = (s==Rational(3,4))"},
{"claim":"Q4c need 99 terms for 1/(N+1)<=0.01","code":"N=99; result = (Rational(1,N+1)<=Rational(1,100)) and (Rational(1,N)>Rational(1,100))"},
{"claim":"Q5b limit equals 1","code":"x=symbols('x'); L=limit((exp(x)-1-x)/(1-cos(x)),x,0); result = (L==1)"},
{"claim":"Q6 radius R=3","code":"x=symbols('x'); r=limit((x/(x+1))*Rational(1,3),x,oo); result = (Rational(1,3)*3==1) and (r==Rational(1,3))"}
]