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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?

Saaqib Mahmood
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    You can evaluate the limit. See here: https://math.stackexchange.com/questions/2682088/show-that-a-n-sum-limits-k-n12n1-frac1k-converges-to-frac12/2682104#2682104 – farruhota Mar 23 '18 at 11:38

3 Answers3

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Your proof is correct.

It is too early (in the book) to obtain the limit. Some time later it will be easy to see an integral sum of $$ \int_{0}^{1}\frac{1}{1+x}\,dx. $$

Przemysław Scherwentke
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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 $$\int_0^1\frac{1}{x+1}=\lim_{n\rightarrow\infty}\sum_{i=1}^n\frac{1}{n}f(\frac{i}{n})=\sum_{i=1}^n\frac{1}{n+i}=\log(x+1)|_{x=0}^{x=1}=\log 2$$

Américo Tavares
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Xin Fu
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What you did is fine.

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.

José Carlos Santos
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