Question

1. Test the following Series for convergence:
1÷1.2 +2÷3.4+3÷5.6​

Answer 1

Answer:

The nth term is n/(2n - 1 + n/5) = 1/(2 - 1/n + 1/5) = 1/(11/5 - 1/n).

The limit of |(n+1)th term|/|nth term| as n -> inf is given by...

lim (11/5 - 1/n)/(11/5 - 1/(n+1)) > lim (11/5)/(11/5 - 1/n) = lim 11/(11 - 5/n) = 1.

Thus, by the ratio test, the series diverges (to +inf).

Alternatively, note that 1/(11/5 - 1/n) > 1/(1 - 1/n) = 1/((n-1)/n) = n/(n-1).

Since lim n/(n-1) = 1, this (other) series diverges. The divergence of our series then follows via the comparison test.

Most simply, we can note that lim 1/(11/5 - 1/n) = 5/11 > 0. This is a sufficient condition for the series to diverge.

I think it will help u

Answer 2

Answer:

Step-by-step explanation:

I assume that this is multiplication which come normarlly in series questions

. = *

Lets start

Nth term is = n / (2n- 1) * 2n

Lim n tends to infinity n / ( 2n -1) *2n

= 1/2n-1

Take n as common

Un = 1/n (2 - 1/n )

Take Vn = 1/n

Lim n tends to infinity Un/Vn = 1/n( 2 -1/n ) /1/n

Un/Vn = 1 / (2 - 1/n ) .: 1/n= 0 due to n tends to infinity.

Un/ Vn =1/2

Hence both converge or diverge together.

Vn = 1/n

n = 1 by comparison test it p <or = 1 is divergent

Hence both series are divergent by comparison test.