Definite Integral as Limit of a sum | ISI QMS | QMA 2019

Find the value of : $ \displaystyle \lim_{n \to \infty}\big[\frac1n+\frac{1}{n+1}+\ldots + \frac{1}{3n}\big]$

Putting it into standard form :

$\displaystyle\lim_{n \to \infty} \big[\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{3n}\big]$

$= \displaystyle \lim_{n\to \infty}\big[\frac{1}{n+0}+\frac{1}{n+1}+\ldots+\frac{1}{n+2n}\big]$

$= \displaystyle \lim_{n \to \infty} \displaystyle\sum_{r=0}^{2n} \frac{1}{n+r}$

$= \displaystyle \lim_{n \to \infty} \frac1n\displaystyle\sum_{r=0}^{2n} \frac{1}{1+\frac rn}$

$= \displaystyle \lim_{n \to \infty} \frac1n\displaystyle\sum_{r=0}^{2n} f(\frac rn)$ where $f(x)=\frac{1}{1+x}$

An useful result : Let $f:[a,b]\to \mathbb R$ be integrable on $[a,b]$ and $\{P_n\}$ be a sequence of partitions of $[a,b]$ such that the sequence $\{\parallel P_n\parallel\}$ converges to $0$. Then if $\epsilon >0$ be given, there exists a natural number $k$ such that $|S(P_n, f)-\int_a^b f|<\epsilon \quad \forall n\geq k$ where $S(P,f)$ is a Riemann sum of $f$ corresponding to $P$ and any choice of intermediate points.

i.e., If $f$ be integrable on $[a,b]$ and $\{P_n\}$ be a sequence of partitions of $[a,b]$ such that $\lim\limits_{n \to \infty} \parallel P_n \parallel=0$, then $\lim\limits_{n\to \infty} S(P_n,f)=\int_a^b f$

Let $P_n=(0,\frac1n,\frac2n,\ldots,\frac{2n}{n})$ be a sequence of partition on $[0,2]$ dividing it into $2n$ sub-intervals of equal length $\frac1n$.

Also $\lim \parallel P_n\parallel=\lim \frac1n=0$.

Let us choose $\xi_r=\frac rn,\quad r=1,2,3,\ldots,2n$

Then the Riemann sum for $f$ on the interval $[0,2]$ corresponding to the partition $P_n$ and chosen intermediate points $\xi_r$

$S(P_n,f)=\frac1n\displaystyle\sum_{r=1}^{2n} f(\frac rn)$

As $f$ is continuous on $[0,2]$, $f$ is integrable on $[0,2]$.

Now can you use the above result to reach to the solution ?

Using the result

$\lim\limits_{n\to\infty} \frac1n\displaystyle\sum_{r=1}^{2n} f(\frac rn) = \displaystyle\int_0^2 f(x) \mathrm d x$

$= \displaystyle\int_0^2 \frac{ \mathrm d x }{1+x} $

$=\ln(1+x)\bigg|_0^2= \ln 3$ [Ans]