Are-n-1-1-4n-2-1-and-n-1-1-n-n-1-n-2-convergent-

Question Number 182940 by mathocean1 last updated on 17/Dec/22

A r e \sum_{n ⩾ 1} \frac{1}{4 n^{2} - 1} a n d

\sum_{n ⩾ 1} \frac{1}{n (n + 1) (n + 2)} c o n v e r g e n t ?

Answered by JDamian last updated on 17/Dec/22

a_{n} = \frac{1}{4 n^{2} - 1} = \frac{1}{(2 n - 1) (2 n + 1)} =

= \frac{1}{2} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1}) t e l e s c o p i n g s e r i e s

S = \frac{1}{2} \underset{k = 1}{\sum^{\infty}} (\frac{1}{2 n - 1} - \frac{1}{2 n + 1}) =

= \frac{1}{2} (\frac{1}{1} - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \dots) =

= \frac{1}{2} [1 - \lim_{n \to \infty} (\frac{1}{2 n + 1})] = \frac{1}{2}

Answered by ARUNG_Brandon_MBU last updated on 17/Dec/22

S = \sum_{n ⩾ 1} \frac{1}{n (n + 1) (n + 2)} = \sum_{n ⩾ 1} (\frac{1}{2 n} - \frac{1}{n + 1} + \frac{1}{2 (n + 2)})

= \frac{1}{2} \sum_{n ⩾ 1} \frac{1}{n} - (\sum_{n ⩾ 1} \frac{1}{n} - 1) + \frac{1}{2} (\sum_{n ⩾ 1} \frac{1}{n} - 1 - \frac{1}{2})

= 1 - \frac{1}{2} (1 + \frac{1}{2}) = 1 - \frac{3}{4} = \frac{1}{4}