How do I prove that the limit of the sequence exists: $a_n = \frac1n +\frac 1{n+1} + \frac1{n+2} + \dots + \frac1{2n}$? How do I prove that the limit of the sequence exists: $$a_n = \frac1n +\frac 1{n+1} + \frac1{n+2} + \dots + \frac1{2n}$$
I can find the limit as n goes to infinity, $\frac1{2n}$ goes to $0$, so the limit is $0$ right? But how do I prove that the limit exists? Do I have to use the $\varepsilon$ $\delta$ definition of a limit?  
For part b) I have to show that the limit is less then $1$ but not less then $\frac12$, I thought the limit was $0$?
 A: We deal with existence of the limit. Note that the $a_k$ are positive, Now we will show that $a_{n+1}\lt a_n$ for all $n$. This is straightforward, for
$$a_{n+1}=a_n+\frac{1}{2n+1}+\frac{1}{2n+2}-\frac{1}{n}\lt a_n,$$
because $\frac{1}{2n+1}+\frac{1}{2n+2}\lt \frac{2}{2n+1}\lt \frac{1}{n}$.
To finish, recall that a decreasing sequence which is bounded below has a limit.
A: Another point of view:
If you know about integration and Riemann sums, you can write your expression as
$$
\frac{1}{n}\frac{1}{1+0/n}+\frac{1}{n}\frac{1}{1+1/n}+\frac{1}{n}\frac{1}{1+2/n}+\cdots+\frac{1}{n}\frac{1}{1+(n-1)/n}+\frac{1}{n}\frac{1}{1+n/n},
$$
which is a Riemann sum for
$$
\int_0^1\frac{1}{1+x}\,dx=\bigl[\log(1+x)\bigr]_0^1=\log2-\log1=\log2.
$$
Since the function $x\mapsto 1/(1+x)$ is continuous in $[0,1]$, we find that, as $n\to+\infty$,
$$
\frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\to \int_0^1\frac{1}{1+x}\,dx=\log2\approx 0.69.
$$
A: It appears from your question that you’ve misunderstood what you’re supposed to do. To make it clearer, sit down and calculate a few terms of your sequence:
$$\begin{align*}
a_1&=\frac11+\frac12=\frac32\\
a_2&=\frac12+\frac13+\frac14=\frac{13}{12}\\
a_3&=\frac13+\frac14+\frac15+\frac16=\frac{19}{20}\\
a_4&=\frac14+\frac15+\frac16+\frac17+\frac18=\frac{743}{840}
\end{align*}$$
The fact that the numbers $\frac1{2n}$ approach $0$ tells you nothing much about what the sequence $\langle a_1,a_2,a_3,\ldots\rangle$ is doing.
You might notice that $a_1>a_2>a_3>a_4$. That suggests trying to show that the sequence is decreasing, meaning that $a_{n+1}<a_n$ for each $n$. The easiest way to do that is to show that $a_n-a_{n+1}>0$. What is $a_n-a_{n+1}$? It’s
$$\left(\frac1n+\frac1{n+1}+\ldots+\frac1{2n}\right)-\left(\frac1{n+1}+\ldots+\frac1{2n}+\frac1{2n+1}+\frac1{2n+2}\right)\;.\tag{1}$$
Now do a lot of cancelling and express the result as a single fraction; it will be easy to check that this fraction is positive.
Since the sequence is decreasing, and $a_3$ is clearly less than $1$, any limit must certainly be less than $1$. To show that the limit is not less than $\frac12$, think about the following questions:


*

*What is the smallest term in the sum for $a_n$?  

*How many terms are there in the sum for $a_n$?

