Problem in finding the value of this limit

Question

Solve the following limit:

$\lim_{n\rightarrow\infty}\frac{1}{n}\left (\frac{n}{n+1}+\frac{n}{n+2}+...+\frac{n}{n+n} \right )$

my approach

I know this problem has been asked before. But my doubt is why I am getting two different answer by using two different methods. First, if I want to solve the given problem using the Cauchy first theorem of limits , then I can consider:

$a_n=\frac{n}{n+k}$.

Clearly, $a_n\rightarrow1$. So, being the average of $n$ such terms, our limit should also tend to $1$(according to Cauchy first theorem of limit). Now, if I want to use the integral approach, I can write the above limit as:

$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{r=0}^{n}\frac{1}{1+\frac{r}{n}}$,

The above mentioned sum, then can be easily solved as:

$\int_{0}^{1}\frac{1}{1+x}dx=ln2$

I don't understand what is going on here. How can we get two different answers for the same problem. Maybe, I am making any mistake or something in my steps is wrong. Or I have misread the Cauchy theorem. It will be very beneficial for me if this doubt is tackled. Thanks.

Answer 1

"$a_n= \frac{n}{n+k}$, clearly $a_n\rightarrow 1$" ??

I didn't know that clearly $\frac{n}{n+n}\rightarrow 1$, $\frac{n}{n+(n-1)}\rightarrow 1$, ....

In the first "method", you are using an "$\lim_{n\rightarrow+\infty}\lim_{k\rightarrow+\infty}c_{n,k}=\lim_{k\rightarrow+\infty}\lim_{n\rightarrow+\infty}c_{n,k}$" argument, which is valid only if the convergence is uniform with respect to one of the indices...

If you do not verify this hypothesis, you will have the same problem with $$ c_{n,k} :=\frac{n}{n+k}$$ $\lim_{n\rightarrow+\infty}\lim_{k\rightarrow+\infty}c_{n,k}=\lim_{n\rightarrow+\infty} 0= 0$, but $\lim_{k\rightarrow+\infty}\lim_{n\rightarrow+\infty}c_{n,k}=1$... (which is by the way a simpler version of the sequel you considered)...

Answer 2

You are assuming that you are allowed to exchange a $\lim$ and a $\sum$, but that is not always the case. For instance, consider the sequence of functions $f_n(x) = e^{-(x-n)^2}$ defined over $\mathbb{R}^+$. It is pretty clear that for any $x\in\mathbb{R}^+$ we have $\lim_{n\to +\infty} f_n(x)=0$, but

$$ \sqrt{\pi}=\lim_{n\to +\infty}\int_{0}^{+\infty}f_n(x)\,dx \color{red}{\neq } \int_{0}^{+\infty}\lim_{n\to +\infty}f_n(x)\,dx = 0.$$

The main theorem regulating the cases in which you are allowed to exchange two operators among $\lim,\int,\sum$ is the dominated convergence theorem.

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Anyway, the original problem can be solved in a elementary fashion, not involving any exchange of operators or any Riemann sum. We have $$ \frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n}=H_{2n}-H_n $$ and since $H_n = \log(n)+\gamma+o(1)$ the limit of the above sum as $n\to +\infty$ is clearly $\log 2$.
The same can be proved by invoking the Hermite-Hadamard inequality ($\frac{1}{x}$ is convex on $\mathbb{R}^+$) and squeezing.