series problems

problem set 5

1

$a_0=1,a_1=-\frac{1}{\sqrt{3}},a_2=\frac{1}{3},a_3=-\frac{1}{3\sqrt{3}}$

$b_n={\sum }_{k=1}^n{a}_k$
$B={\lim }_{n\to \infty }{b}_n$
$|b_n-B|<ϵ$, i.e. sum from $n$ to $\infty$

$b_n={\sum }_{k=1}^n(-\frac{1}{\sqrt{3}}{)}^k$
$r=-\frac{1}{\sqrt{3}}$, sum is $\frac{1-r^n}{1-r}$
$n=\infty$, sum is $\frac{1}{1-r}$

$|b_n-B|<ϵ$
$|\frac{1-r^n}{1-r}-\frac{1}{1-r}|<ϵ$
$|-r^n|<ϵ\cdot (1-r)$
solve for $n$ for $M$?

7

${\sum }_{n=1}^{\infty }\frac{\sqrt{n+1}-\sqrt{n-1}}{2}$

$\frac{\sqrt{2}-\sqrt{0}}{2},\frac{\sqrt{3}-\sqrt{1}}{2},\frac{\sqrt{4}-\sqrt{2}}{2},\frac{\sqrt{5}-\sqrt{3}}{2}$
partials: $0.707,1.073,1.366,1.618$

$\frac{1}{2}({\sum }_{n=1}^{\infty }\sqrt{n+1}-\sqrt{n-1})$
$\frac{1}{2}[(\sqrt{2}-\sqrt{0})+(\sqrt{3}-\sqrt{1})+(\sqrt{4}-\sqrt{2})+(\sqrt{5}-\sqrt{3})]$
$\frac{1}{2}[(\sqrt{2}-\sqrt{2})+(\sqrt{3}-\sqrt{3})-\sqrt{0}-\sqrt{1}+\sqrt{4}+\sqrt{5}]$

so $-\sqrt{1}$ (i.e. $1$), $\sqrt{4}$, $\sqrt{5}$ don’t cancel
$\sqrt{4}$ and $\sqrt{5}$ generalize to $\sqrt{N}$ and $\sqrt{N+1}$
so $S=li{m}_{n\to \infty }[-1+\sqrt{n}+\sqrt{n+1}]\times \frac{1}{2}$
$S=li{m}_{n\to \infty }[\sqrt{n}+\sqrt{n+1}]\times \frac{1}{2}=\infty$

slides

${\sum }_{n=3}^{\infty }\frac{(-1{)}^{n}n}{(n+1)(2-n)}$
$(-1{)}^n\cdot \frac{n}{(n+1)(2-n)}$

${\lim }_{n\to \infty }\frac{n}{n+2-n^2}$
${\lim }_{n\to \infty }\frac{n}{n+2-n^2}\cdot \frac{\frac{1}{n^2}}{\frac{1}{n^2}}$
${\lim }_{n\to \infty }\frac{\frac{1}{n}}{\frac{1}{n}+\frac{2}{n^2}-1}$
$\frac{0}{0+0-1}=\frac{0}{-1}=0$
converges at $0$ at $\infty$

$b_n>b_{n+1}>0$
$\frac{n}{n+2-n^2}>\frac{n+1}{n+1+2-(n+1{)}^2}$
$\frac{n+1}{n+1+2-(n+1{)}^2}=\frac{n+1}{n+3-n^2-1-2n}=\frac{n+1}{-n+2-n^2}$
$\frac{n}{n+2-n^2}>\frac{n+1}{-n+2-n^2}$, inequality holds