Exercise.

Consider the sequence defined by ${a}_{1}=1$ and ${a}_{n+1}=\sqrt{1+{a}_{n}}$.

(a) Assuming for the moment that $\left({a}_{n}\right)$ is
convergent, by solving a quadratic equation or otherwise
make a reasonable conjecture

for a value $l$
that might be the limit of the sequence.

(b) Prove that $1\u2a7d{a}_{n}<l$ holds for all $n$, by induction. Your proof must not use the conjecture from (a) that ${a}_{n}\to l$, only the value of $l$ itself.

(c) Using part (b), or by induction, prove that the sequence is monotonic nondecreasing.

(d) Which theorem from lectures allows you to deduce from (b) and (c) that ${a}_{n}$ has a limit?

(e) Apply the continuity of the function $f\left(x\right)=\sqrt{1+x}$ (considered as a function of positive $x$ into the set of positive reals) to prove that the limit of the sequence $\left({a}_{n}\right)$ is $l$, as conjectured in part (a).

Exercise.

Find $supA$ and $infA$ where $A=\{\frac{1}{n}-\frac{1}{k}:n,k\in \mathbb{N},1<n<k\}$.

Exercise.

For the following series series $\sum _{n=1}^{\infty}{a}_{n}$, determine if the series converges and if so find the limit. (Hint: use partial fractions to express ${a}_{n}$.)

(a) ${a}_{n}=\frac{1}{n(n+1)}$

(b) ${a}_{n}=\frac{1}{{n}^{2}+2n}$

(c) ${a}_{n}=\frac{1}{n({n}^{2}-1)}$ (As ${a}_{1}$ is not defined consider $\sum _{n=2}^{\infty}{a}_{n}$.)

Exercise.

(a) Let $p>0$ in $\mathbb{R}$ and $n\in \mathbb{N}$. Prove that $(1+p{)}^{n}\u2a7e1+np+\frac{n(n-1)}{2}{p}^{2}$. (Use induction on $n$.)

(b) If $r\in \mathbb{R}$ with $0<r<1$, show that $n{r}^{n}\to 0$ as $n\to \infty $. (Hint: write $r=\frac{1}{1+p}$ and use (a).)

(c) Let ${a}_{n}=n{r}^{n}$ where $0<r<1$. Show that $\sum _{k=1}^{n}{a}_{k}=\frac{r}{(1-r{)}^{2}}(1-(n+1){r}^{n}+n{r}^{n+1})$ (Use induction on $n$.)

(d) Hence show that the series $\sum _{n=1}^{\infty}n{r}^{n}$ converges if $0<r<1$, and find its limit.

(e) Show that $\sum _{n=1}^{\infty}n{r}^{n}$ diverges if $r\u2a7e1$.

Standard results and theorems, including the squeeze rule, may be used if quoted correctly.

Let ${s}_{n}=\sum _{k=1}^{n}{a}_{n}$.

Note that ${a}_{n}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ so ${s}_{n}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots +\frac{1}{n}-\frac{1}{n+1}$ hence ${s}_{n}=\frac{1}{1}-\frac{1}{n+1}$ for all $n$. But $\frac{1}{n+1}\to 0$ as $n\to \infty $ (being a subsequence of a standard null sequence $\frac{1}{n}$) and so ${s}_{n}\to 1-0=1$ by continuity of $-$ and therefore the series converges with $\sum _{n=1}^{\infty}{a}_{n}=1$.

Since $0<r<1$ there is a real number $p$ such that $r=\frac{1}{1+p}$ and $p>0$. Now note that $0\u2a7dn{r}^{n}=\frac{n}{(1+p{)}^{n}}\u2a7d\frac{n}{1+np+\frac{n(n-1)}{2}{p}^{2}}$ by part (a). Simplifying this we have $0\u2a7dn{r}^{n}\u2a7d\frac{\frac{2}{n}}{\frac{2}{{n}^{2}}+2\frac{p}{n}+{p}^{2}-\frac{{p}^{2}}{n}}$. The sequence $\left({b}_{n}\right)$ on the right hand side of this inequality converges to $0$ by the standard null sequence $\frac{1}{n}$ and continuity of addition, subtraction, multiplication, and of division at $(0,{p}^{2})$. (Note that ${p}^{2}\ne 0$.) Thus by the squeeze rule applied to $0\u2a7dn{r}^{n}\u2a7d{b}_{n}\to 0$ we have $n{r}^{n}\to 0$ as $n\to \infty $ as required.