Test the convergence of the series 1+1/1.2+1/2.2^2+1/2.2^3+...

We have to FIND the convergence of the infinite series1 + 1/2² + 2²/3³ + 3³/4⁴ + 4⁴/5⁵ +…...………….(1)Since taking AWAY a part of an infinite series does not affect its behaviour, we can study the behaviour of the derived series2²/3³ + 3³/4⁴ + 4⁴/5⁵ +…...………………………(2)so that whatever is true for (2) is true for (1). The series (2) is monotonically decreasing as every term of the series is smaller than its preceding term.nth term u(n) = (n+1)^(n+1)/(n+2)^(n+2)= [n^(n+1).(1+1/n)^(n+1)]/[n^(n+2).(1+2/n)^(n+2)]=(1/n) . [(1+1/n)^(n+1)/(1+2/n)^(n+2)]Now as n→ ∞, 1+1/n→1+0 = 1 and 1+2/n→1+0 = 1. As a CONSEQUENCE,(1+1/n)^(n+1)/(1+2/n)^(n+2)→1 for n very large.But 1/n→0 as n→ ∞.∴ the nth term of series (2)→0 .∴ the infinite series (2) is convergent and hence the original series (1) is convergent.Let us compute the value of each term of the series (1).1 = 11/2² = 1/4 = .252²/3³ = 4/27 = .1483³/4⁴ = 27/256 = .10554⁴/5⁵ = 256/3125 = .0819