$x \lim_{n\to\infty}\sum^n_{r=1}\frac{1}{n}\ln\left(\frac{1+\frac{xr}{n}}{1+\frac{x²r²}{n²}}\right)$ into an integral.

Question

$$x \lim_{n\to\infty}\sum^n_{r=1}\frac{1}{n}\ln\left(\frac{1+\frac{xr}{n}}{1+\frac{x²r²}{n²}}\right)$$ into an integral.

So this beast right here is the result of a bigger one: $$\lim_{n\to\infty}\left(\frac{n^n(x+n)(x+\frac{x}{2})\dots (x+\frac{n}{n})}{n!(x²+n²)(x²+\frac{n²}{4})\dots (x²+\frac{n²}{n²})}\right)^{\frac{x}{n}}$$ but that's another thing, no need to check on that since I know for a fact that it is correct and it does arrive to the thing above. All I have left os to transform that riemanm sum into a definite integral, but the issue is that I'm unsure on how to do so. Any hints or tips? I've noticed that $dx=\frac{1}{n}$, since it's outside of the function $\ln$ but that's the most I have been able to get to.

Answer 1

An overview

• For simplicity, suppose $f:[a,b]\rightarrow\mathbb{R}$ is a nice continuous function.
• Divide the interval in $n$ pieces, not necessarily of the same length. That is, consider points $\mathcal{P}=\{a=t_0<t_1<\ldots<t_n=b\}$.
• For each subinterval $I_k=[t_{k-1},t_k]$, ($k=1,\ldots,n$), choose a point $t^*_k\in I_k$. Call this $t^*_k$'s tags, and denote the set of tags as $\tau_{\mathcal{P}}$.
• The Riemann sum for the partition $\mathcal{P}$ and tans $\tau_{\mathcal{P}}$ is defined as $$S(f,\mathcal{P},\tau_{\mathcal{P}})=\sum^n_{k=1}f(t^*_k)(x_k-x_{k-1})$$
• As you may have seen from your Calculus course, as the partition $\mathcal{P}$ gets finer and finer, and no matter what tags $\tau_{\mathcal{P}}$ one chooses the Riemann sums converge to a number, denoted as $\int^b_a f$ (or $\int^b_b f(t)\,dt$).
• Equivalently, (all this takes substantial effort to make it rigorous terms but let's not worry about that here) as the size of the partition $\|\mathcal{P}\|=\max_{1\leq k\leq n}(x_k-x_{k-1})$ gets smaller and smaller, no matter what tags $\tau_{\mathcal{P}}$ one chooses, the Riemann sums converge to a number, denoted as $\int^b_a f$ (or $\int^b_b f(t)\,dt$).

So far I am probably not telling anything new, but let's use this as a simple overview.

As you may know already, all continuous functions on $[a,b]$ are Riemann integrable, that is the limit of the Riemann sums exists. In particular, we may consider uniform partitions of $[a,b]$, that is partitions of the form $t_k=a+\frac{b-a}{n}k$, $k=0,\ldots, n$, in which case $x_k-x_{k-1}=\frac{b-a}{n}=h_n$. Since the choice of tags is not important, we may as well choose the endpoints (right for example) of the subintervals $I_k=\big[a+h_n(k-1),a+h_nk]$, $k=1,\ldots,n$. Then $$\lim_{n\rightarrow\infty}\sum^n_{k=1}f\big(a+k\tfrac{b-a}{n}\big)\frac{b-a}{n} =\lim_{n\rightarrow\infty} h_n\sum^n_{k=1}f(a+kh_n)=\int^b_a f$$

Back to you original problem: The sum in your problem can be written as $$ S_n=x\sum^n_{k=1}\frac{1}{n}\ln\Big(\frac{1+\tfrac{xk}{n}}{1+\big(\tfrac{xk}{n}\big)^2}\Big)=\frac{x}{h}\sum^n_{k=1}\ln\Big(\frac{1+\tfrac{x}{n}k}{1+\big(\tfrac{x}{n}k\big)^2}\Big)= h_n\sum^n_{k=1}\ln\Big(\frac{1+h_nk}{1+\big(h_nk\big)^2}\Big) $$ where $h_n=\frac{x}{n}$. This corresponds to a Riemann sum with uniform partition over $[0,x]$ with $n$ subintervals of size $h_n$ and using the right-end points of subintervals as tags, and function $f(t)=\ln\Big(\frac{1+t}{1+t^2}\Big)$. Hence $$S_n\xrightarrow{n\rightarrow\infty}\int^x_0\ln\Big(\frac{1+t}{1+t^2}\Big)\,dt$$

I hope this clarify matters.