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 $0$?

Solution 1:

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 $n\to+\infty$, $$ \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{2n}\to \int_0^1\frac{1}{1+x}\,dx=\log2\approx 0.69. $$

Solution 2:

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.