Question

The sum of the series
$\frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} ........ \infty$
is
$a) \frac{1}{6}$
$b) \frac{1}{3}$
$c) \frac{1}{2}$
$d) \frac{5}{6}$



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Answer 1

Answer:

1/6

Step-by-step explanation:

$rth \: term = \frac{1}{(2r + 1)(2r + 3)}$

i have used su(..) to denote summation!!!!!

Sn=su(rth term)

Sn=su(1/(2r+1)(2r+3))

Sn=su(1/2 * (2r+3)-(2r+1)/(2r+1)(2r+3) )

Sn=su(1/2 * { 1/(2r+1) - 1/(2r+3)} )

Sn=1/2 *( (1/3-1/5) + (1/5-1/7) +......+(1/(2n+1) -1/(2n+3))

Sn=1/2 *(1/3 -1/(2n+3))

[↑as the rest of the terms cancel out]

Sn=1/2* (2n/3(2n+3))

Sn=n/3(2n+3)

now as n→∞ then ,1/n→0...

Lim Sn

n→∞

=>Lim n/(3(2n+3))

n→∞

=>Lim n * 1/3n * 1/(2+3/n)

n→∞

=>Lim 1/3 * 1/(2+3*1/n)

n→∞

now when n→∞ then ,1/n→0...

=>1/3 * 1/2

=>1/6 (answer)