Question

$\displaystyle \rm \lim_{n \to \infty } \sum_{r = 0}^{n(b - a)} \frac{1}{na + r}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\displaystyle \rm \lim_{n \to \infty } \sum_{r = 0}^{n(b - a)} \frac{1}{na + r}$

can be rewritten as

$\rm \: =  \:\displaystyle \rm \lim_{n \to \infty } \: \frac{1}{na} + \displaystyle \rm \lim_{n \to \infty } \sum_{r = 1}^{n(b - a)} \frac{1}{na + r}$

$\rm \: =  \:0 + \displaystyle \rm \lim_{n \to \infty } \sum_{r = 1}^{n(b - a)} \frac{1}{na + r}$

$\rm \:=  \: \displaystyle \rm \lim_{n \to \infty } \sum_{r = 1}^{n(b - a)} \frac{1}{na + r}$

can be further rewritten as

$\rm \: =  \:\displaystyle \rm \lim_{n \to \infty } \sum_{r = 1}^{n(b - a)} \frac{1}{n\bigg(a + \dfrac{r}{n}\bigg) }$

can be further rewritten as

$\rm \: =  \:\displaystyle \rm \lim_{n \to \infty } \: \frac{1}{n} \: \sum_{r = 1}^{n(b - a)} \frac{1}{a + \dfrac{r}{n} }$

Now, using Limit as a Sum of Definite integrals, we have

$\rm :\longmapsto\:\dfrac{r}{n} \: changes \: to \: x$

$\rm :\longmapsto\:\dfrac{1}{n} \: changes \: to \: dx$

$\rm :\longmapsto\:\displaystyle \sf \lim_{n \to \infty} \sum \: changes \: to \: \int$

$\rm :\longmapsto\: \: lower \: limit \: a \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{1}{n} = 0$

$\rm :\longmapsto\: \: upper \: limit \: b \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{n(b - a)}{n} = \: b - a$

So, given expression can be rewritten as

$\rm \: =  \:\displaystyle\int_0^{b-a}\rm \: \frac{1}{a + x} \: dx \:$

$\rm \: =  \:\bigg[log |x + a|\bigg] _0^{b-a}$

$\rm \: =  \:log |b - a + a| - log |a + 0|$

$\rm \: =  \:logb \: - \: loga$

$\rm \: =  \:log\bigg |\dfrac{b}{a} \bigg|$

Hence,

$\rm\implies \:\boxed{\sf{  \:\rm \: \displaystyle \rm \lim_{n \to \infty } \sum_{r = 0}^{n(b - a)} \frac{1}{na + r}=  \:log\bigg |\dfrac{b}{a} \bigg| \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$