Question

find the sum of the infinity of the series 1/2.4+1/4.6+1/6.8+1/8.10

Answer 1

S=1/4(1.2+1/2.3+1/3.4+1/4.5+........)
=1/4(1-1/2)+(1/2-1/3)+(1/3-1/4)+......+(1/n-1/n+1)
=1/4(1-1/n+1)
=1/4(n/n+1)

Answer 2

S = 1/2.4 + 1/4.6 + 1/6.8 + 1/8.10 + .....

first of all, we have to find nth term,
see first number of denominator, e.g., 2, 4, 6, 8, 10 .... so, nth term of this series = 2n
and 2nd number of denominator , e.g., 4, 6 , 8, 10, .... so, nth term of this series = (2n + 2)

hence, nth term of given sequence is 1/{2n(2n+2)}

so, $S=\displaystyle\Sigma^{n=r}_{n=1}{\frac{1}{(2n)(2n+2)}}$

let's break Tn = 1/(2n)(2n+1) = 1/4[ 1/n(n+1)]
= 1/4[1/n - 1/(n + 1)]

when n = 1 , T1 = 1/4 [ 1/1 - 1/2 ]

n = 2 , T2 = 1/4 [1/2 - 1/3 ]

n = 3, T3 = 1/4 [ 1/3 - 1/4 ]

........... ..... ............
.............. ......... ..
n = r , Tr = 1/4 [1/r - 1/(r + 1)]

-------------------------------------
S = T1 + T2 + T3 + T4 ..... + Tr

= 1/4[1/1 - 1/2] + 1/4[1/2 - 1/3] + 1/4[1/3 - 1/4] +..... + 1/4[1/r - 1/(r + 1)]

= 1/4[ 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ..... 1/r - 1/(r + 1)]

= 1/4 [ 1 - 1/(r + 1)]

so, S = 1/4[1 - 1/(r + 1)]

if r tends to infinity then, S = 1/4[1 - 1/(∞ + 1)]
S = 1/4 [ 1 - 0 ] = 1/4

hence, Sum of infinity of the given series is 1/4