Question

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

Answer 1

Answer:

We have to find the convergence of the infinite series

1 + 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 series

2²/3³ + 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 = 1

1/2² = 1/4 = .25

2²/3³ = 4/27 = .148

3³/4⁴ = 27/256 = .1055

4⁴/5⁵ = 256/3125 = .0819