# Analysis help. Let An = 1/(n+1) + 1/(n+2) + ... + 1/(2n), n belonging to the Natural Numbers.?

Let An = 1/(n+1) + 1/(n+2) + ... + 1/(2n), n belonging to the Natural Numbers.

a.) Show that {An} from 1 to infinity is increasing.

b.) Is {An} from 1 to infinity convergent? If so why, if not why?

Relevance

a) A_n+1 - A_n = 1/(2n+1) + 1/(2n+2) - 1/(n+1)

Since 1/(2n+1) + 1/(2n+2) > 2/(2n+2) = 1/n+1, this will always be positive and the series is increasing.

b) Yes - this converges. It will converge to a number that looks like ln(2n) - ln(n) = ln 2... but proving that is a little more difficult. Thinking...

Ed: The argument below about the difference approaching 0 as n-> infinity is NOT valid. If this was the series of 1/sqrt(n+1) ... 1/sqrt(2n), then this would NOT converge.

I believe the proper analytical argument is that 1/(n+a) ~= ln((n+a)/(n+a-1)) = ln(n+a) - ln (n+a-1) as n approaches infinity, therefore the sum telescopes to ~= ln(2n) - ln(n) = ln 2.

• Anonymous
9 years ago

a) To show that {A(n)} from 1 to infinity is increasing, we need to show that {A(n+1)} is always greater than {A(n)}.

Consider A(n+1): A(n+1) = 1/(n + 2) + ... + 1/(2n) + 1/(2n+1) + 1/(2n+2)

If we now look at A(n+1) - A(n): A(n+1) - A(n) = 1/(2n+1) + 1/(2n+2) - 1/(n+1)

--> 1/(2n+1) + 1/2(n+1) - 1/(n+1) --> 1/(2n+1) - 1/2(n+1) = 1/(2(2n+1)(n+1))

(I leave the algebra to you for this final result...)

However, for n >/= 1, that means A(n+1) - A(n) = 1/(2(2n+1)(n+1) > 0. Therefore {A(n)} from 1 to infinity is increasing. (Additional note that will help answer part (b): The difference will go toward zero as n -> infinity...)

b) {A(n)} from 1 to infinity must be convergent, since in the limit the difference between successive terms approaches 0 as n -> infinity.