Is the sequence convergent

Question

I am to determine if the sequence

$$x_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}+\frac{1}{2n} $$

converges or not?

I'm thinking I if I can determine if the sequence has some kind om bound and if it is either increasing or decreasing? Because then I know from the Monotone convergence theorem that it converges.

But I'm not sure how you would prove this.

Answer 1

The sequence is bounded, since

$$x_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}+\frac{1}{2n} \geq \frac{1}{2n}+\frac{1}{2n} + \cdots + \frac{1}{2n} = \frac12$$

and

$$x_n = \frac{1}{n+1}+\frac{1}{n+2}+...+\frac{1}{2n-1}+\frac{1}{2n} \leq \frac{1}{n+1}+\frac{1}{n+1} + \cdots + \frac{1}{n+1} = \frac{n}{n+1}\leq 1$$

To see whether it is monotonic, look at the sign of

$$x_n - x_{n-1}.$$

The expression $x_n-x_{n-1}$ should have a lot of things cancel out.

Answer 2

Hint:

You may write $$\sum_{k=1}^n\frac{1}{n+k} = \frac{1}{n}\sum_{k=1}^n\frac{1}{1+\frac{k}{n}}\stackrel{n \to \infty}{\longrightarrow} \int_0^1\frac{1}{1+x}\; dx$$

Answer 3

Correct if wrong :

$1/2=n/(2n) <x_n < n/(n+1) <1,$

hence bounded.

$x_n = 1/(n+1)+..........1/(2n);$

$x_{n+1}= 1/(n+2) +.....1/(2n+1)+1/(2n+2).$

Note : $1/(2n+2)= (1/2)(1/(n+1)),$ and

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

$(1/2)(1/(n+1))$.

$1/(2n+1)+1/(2n+2) >$

$1/(n+1).$

Hence

$x_{n+1} > x_n.$

$x_n$ is bounded, increasing , hence convergent.