Question-137403

Question Number 137403 by BHOOPENDRA last updated on 02/Apr/21

Commented by BHOOPENDRA last updated on 02/Apr/21

p l e a s e h e l p m e o u t t h i s i n c o r r e c t w a y

Answered by liberty last updated on 02/Apr/21

\frac{1}{1.3} + \frac{1}{3.5} + \frac{1}{5.7} + \dots = \frac{1}{3} + \underset{n = 1}{\sum^{\infty}} \frac{1}{(2 n + 1) (2 n + 3)}

= \frac{1}{3} + \lim_{n \to \infty} (\frac{1}{3.5} + \frac{1}{5.7} + \frac{1}{7.9} + \dots + \frac{1}{(2 n + 1) (2 n + 3)}

= \frac{1}{3} + \lim_{n \to \infty} \frac{1}{2} [\frac{5 - 3}{3.5} + \frac{7 - 5}{5.7} + \frac{9 - 7}{7.9} + \dots + \frac{(2 n + 3) - (2 n + 1)}{(2 n + 1) (2 n + 3)}]

= \frac{1}{3} + \lim_{n \to \infty} \frac{1}{2} [\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \dots + \frac{1}{2 n + 1} - \frac{1}{2 n + 3}]

= \frac{1}{3} + \frac{1}{2} \lim_{n \to \infty} [\frac{1}{3} - \frac{1}{2 n + 3}]

= \frac{1}{3} + \frac{1}{6} = \frac{1}{2}

Commented by BHOOPENDRA last updated on 02/Apr/21

w h a t a b o u t t h e u p p e r p a r t o f t h e q u e s t i o n s i r

Commented by BHOOPENDRA last updated on 02/Apr/21

y o u h a v e t o e x p r e s s i n I f o r m

Commented by BHOOPENDRA last updated on 02/Apr/21

r e a d t h e q u e s t i o n a g a i n, y o u a r e d e v l o p e

f (x) a s a f o u r i e r s e r i e s a n d t h e n

e v a l u a t e t h i s f (x) = i (x) h e r e