# Prob. 10, Sec. 3.3, in Bartle & Sherbert's INTRO TO REAL ANALYSIS, 4th ed: Is this sequence convergent?

Here is Prob. 10, Sec. 3.3, in the book Introduction to Real Analysis by Robert G. Bartle & Donald R. Sherbert, 4th edition:

Establish the convergence or the divergence of the sequence $\left( y_n \right)$, where $$y_n \colon= \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \ \mbox{ for } n \in \mathbb{N}.$$

My Attempt:

For each $n \in \mathbb{N}$, we have \begin{align} y_{n+1} - y_n &= \left( \frac{1}{ (n + 1) +1} + \frac{1}{ (n+1) +2 } + \cdots + \frac{1}{2(n+1)} \right) - \left( \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \right) \\ &= \left( \frac{1}{ n + 2} + \cdots + \frac{1}{2n+ 2 } \right) - \left( \frac{1}{ n+1} + \cdots + \frac{1}{2n} \right) \\ &= \frac{ 1 }{2n+1} + \frac{ 1}{2n+2} - \frac{1}{n+1} \\ &= \frac{1}{2n+1} - \frac{1}{2n+2} \\ &= \frac{1}{ (2n+1) (2n+2) } > 0, \end{align} which shows that our sequence is monotonically increasing.

Also, for each $n \in \mathbb{N}$, we have $$y_n = \frac{1}{ n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} < \underbrace{\frac{1}{n} + \frac{1}{n} + \cdots + \frac{1}{n} }_{ \mbox{ n times } } = 1.$$

Thus our sequence, being monotonically increasing and bounded, is convergent.

Is what I've done so far all correct?

How to determine the limit of this sequence?

Yes, you are all right. Because an increasing and bounded real number sequence is convergent, it's the axiom of Real number: A bounded set on $R$ has supremum. For calculate the limit, consider function $$f(x)=\frac{1}{x+1}\ \ \text{on}\ \ [0,1]$$it's integrable, and we let $0<\frac{1}{n}<\frac{2}{n}<……<\frac{n-1}{n}<1$, hence $$\int_0^1\frac{1}{x+1}=\lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{1}{n}f(\frac{i}{n})=\sum_{i=1}^n\frac{1}{n+i}=\log(x+1)|_{x=0}^{x=1}=\log 2$$
It is too early (in the book) to obtain the limit. Some time later it will be easy to see an integral sum of $$\int_{0}^{1}\frac{1}{1+x}\,dx.$$
Now, as far as the imit is concerned, note that$$y_n=\frac1n\left(\frac1{1+1/n}+\frac1{1+2/n}+\cdots+\frac1{1+n/n}\right).$$This is a Riemann sum.