Question

\[\lim_{n \rightarrow \infty}(\frac{ 1 }{ 1*3 }+\frac{ 1 }{ 3*5 }+......\frac{ 1 }{ (2n-1)(2n+1) })\]

Answer 1

2=3-1
so, 1/1*3 = 1/2(2/1*3) = 1/2((3-1)/3*1) = 1/2[1-1/3]
do this for every term.
should i do it in latex, or you got it ?

Answer 2

Latex...Plzz

Answer 3

\[\frac{1}{1 \times 3}=\frac{1}{2} \times \frac{2}{1 \times 3}=\frac{1}{2} \times[\frac{3-1}{1 \times 3}]=\frac{1}{2} \times[\frac{1}{1 }-\frac{1}{ 3}]\]
do this for every term.

Answer 4

2nd term = 1/3 -1/5
notice 1/3 will get cancelled, and if you go on, all the terms excepts 1st and last will get cancelled.

Answer 5

Would u Call This Partial Fraction Decomposition?

Answer 6

yes. i would.

Answer 7

it would be n /(2n+1) ryt? @hartnn

Answer 8

yes.

Answer 9

then limit=... ?

Answer 10

1/2

Answer 11

correct.

Answer 12

Thxxx

Answer 13

use the telescopic's principle, it can be
1/2 (1 - 1/(2n+1))
1/2 (2n/(2n+1))
just look the cofficient of n, they are same) :)