Question

Help me by step by step

Answer 1

let's do this by partial sums

$(\frac{1}{5}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{7})+(\frac{1}{7}-\frac{1}{8})+$ $(\frac{1}{8}-\frac{1}{9})+(\frac{1}{9}-\frac{1}{10})+...(\frac{1}{n+1}-\frac{1}{n+2})$ etc
we notice that there are pairs that cancel
$\frac{1}{5}+(-\frac{1}{6}+\frac{1}{6})+(-\frac{1}{7}+\frac{1}{7})+(-\frac{1}{8}+$ $\frac{1}{8})+(-\frac{1}{9}+\frac{1}{9})+(-\frac{1}{10}+...\frac{1}{n+1})+(-\frac{1}{n+2})$
we will then be left with the end terms which are
$\frac{1}{4+1}-\frac{1}{ \infty+2}$
$\frac{1}{ \infty+2}$ basically simplifies to 0 because it is basically 1/infintiy

so we are left with $\frac{1}{4+1}$ or $\frac{1}{5}$

the sum is $\frac{1}{5}$