Question

how much is 1 + 1.3/6 +1.3.5/6.8 .... upto infinity? Please Explain

Answer 1

given by 
(1 * 3 * ... * (2n-1))/(6 * 8 * ... * (2+2n)): 

Note that 
(1 * 3 * ... * (2n-1))/(6 * 8 * ... * (2+2n)) 
= (1 * 3 * ... * (2n-1))/[2^(n-1) * (3 * 4 * ... * (n+1))] 
= (1 * 3 * ... * (2n-1))/[2^(n-1) * (n+1)!/2] 
= [(2n)!/(2 * 4 * ... * (2n))] / [2^(n-2) (n+1) * n!] 
= [(2n)!/(2^n * n!)] / [2^(n-2) (n+1) * n!] 
= (2n)! / [2^(2n-2) (n+1) (n!)^2]. 

By Stirling's formula, n! ~ (n/e)^n √(2πn). 
(That is, the ratio of these quantities is 1 as n tends to infinity.) 

So, (2n)! / [2^(2n-2) (n+1) (n!)^2] 
~ (2n/e)^(2n) √(2π * 2n) / [2^(2n-2) (n+1) * ((n/e)^n √(2πn))^2] 
= 4/(πn(n+1)) 
~ 4/(πn * n) 
= (4/π) * 1/n^2. 

Since Σ(n = 1 to ∞) (4/π) * 1/n^2 converges (being a constant multiple of a convergent p-series), we conclude that the original series also converges by the Limit Comparison Test.