Question

$\displaystyle \rm \lim_{n \to \infty } \frac{1}{n} \left ( \sum _{k = 1}^{n} \frac{n - (k - 1)}{n} \ln \bigg( \frac{k}{n} \bigg) \right)$​

Answer 1

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

Given expression is

$\rm \: \displaystyle \rm \lim_{n \to \infty } \frac{1}{n} \left ( \sum _{k = 1}^{n} \frac{n - (k - 1)}{n} \ln \bigg( \frac{k}{n} \bigg) \right)$

can be rewritten as

$\rm \: = \: \displaystyle \rm \lim_{n \to \infty } \frac{1}{n} \left ( \sum _{k = 1}^{n} \bigg[1 - \dfrac{k}{n} + \frac{1}{n} \bigg] \ln \bigg( \frac{k}{n} \bigg) \right)$

can be further rewritten as

$\rm \: = \: \displaystyle \rm \lim_{n \to \infty } \frac{1}{n} \left ( \sum _{k = 1}^{n} \bigg[1 - \dfrac{k}{n} \bigg] \ln \bigg( \frac{k}{n} \bigg) \right)$

$\bigg[ \because \: \rm \: \dfrac{1}{n } \to \: 0 \: \: as \: \: n \: \to \: \infty \: \bigg]$

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

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

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

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

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

$\rm :\longmapsto\: \: upper \: limit \: b \: = \: \displaystyle \sf \lim_{n \to \infty}\frac{n}{n} = \: 1$

So, above expression can be rewritten as

$\rm \: = \: \displaystyle\int_0^1\rm (1 - x) \: logx \: dx$

Let evaluate first

$\rm \: \: \displaystyle\int\rm (1 - x) \: logx \: dx$

So, using integration by parts, we get

$\rm \: = \:logx\displaystyle\int\rm (1 - x) \: dx \: - \: \displaystyle\int\rm\bigg[\dfrac{d}{dx} logx \: \displaystyle\int\rm \: (1 - x)dx\bigg]dx$

$\rm \: = \:logx\bigg[x - \dfrac{ {x}^{2} }{2} \bigg] - \displaystyle\int\rm \: \frac{1}{x} \times \bigg[x - \dfrac{ {x}^{2} }{2} \bigg]dx$

$\rm \: = \:logx\bigg[x - \dfrac{ {x}^{2} }{2} \bigg] - \displaystyle\int\rm \: \bigg[1 - \dfrac{ {x} }{2} \bigg]dx$

$\rm \: = \:logx\bigg[x - \dfrac{ {x}^{2} }{2} \bigg] - \: \bigg[x - \dfrac{ {x}^{2} }{4} \bigg]$

Hence,

$\rm \: \: \displaystyle\int_0^1\rm (1 - x) \: logx \: dx$

$\rm \: = \:logx\bigg[x - \dfrac{ {x}^{2} }{2} \bigg] \bigg |_0^1 - \: \bigg[x - \dfrac{ {x}^{2} }{4} \bigg]\bigg |_0^1$

$\rm \: = \:(log1 - 0) - \bigg(1 - \dfrac{1}{4} \bigg)$

$\rm \: = \:0 - \dfrac{3}{4}$

$\rm \: = \:- \: \dfrac{3}{4}$

Hence,

$\\ \boxed{\sf{  \:\displaystyle \rm \lim_{n \to \infty } \frac{1}{n} \left ( \sum _{k = 1}^{n} \frac{n - (k - 1)}{n} \ln \bigg( \frac{k}{n} \bigg) \right) = - \frac{3}{4} \: }}$

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

Formulae Used :-

$\boxed{\sf{  \:\displaystyle\int\rm \: uvdx \: = \: u\displaystyle\int\rm \: vdx - \displaystyle\int\rm\bigg[\dfrac{d}{dx} u\displaystyle\int\rm \: vdx\bigg]dx \: }}$

$\boxed{\sf{  \:\displaystyle\int\rm \: {x}^{n}dx \: = \: \frac{ {x}^{n + 1} }{n + 1} + c \: }}$

$\boxed{\sf{  \:\displaystyle\int\rm \: k \: dx \: = \: kx \: + \: c \: \: }}$

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

Additional Information :-

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

Answer 2

Answer:

$The \: answer \: is \: \dfrac{53}{6}$

Step-by-step explanation:

$\displaystyle \lim_{n\to \infty}\sum_{i=1}^n\dfrac{(2+\frac{i}{n})^2+(2+\frac{i}{n})}{n}$

$\begin{gathered}=\displaystyle \lim_{n\to \infty}\sum_{i=1}^n\dfrac{4+\frac{4i}{n}+\frac{i^2}{n^2}+(2+\frac{i}{n})}{n}\\\\=\displaystyle \lim_{n\to \infty}\sum_{i=1}^n\dfrac{6+\frac{5i}{n}+\frac{i^2}{n^2}}{n}\\\\=\displaystyle \lim_{n\to \infty}\sum_{i=1}^n\dfrac{6n^2+5ni+i^2}{n^3}\\\\=\displaystyle \lim_{n\to \infty}\dfrac{1}{n^3} (6n^2\sum_{i=1}^n1+5n\sum_{i=1}^ni+\sum_{i=1}^ni^2)\\\\=\displaystyle \lim_{n\to \infty}\dfrac{1}{n^3} (6n^2(n)+5n\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6})\\\end{gathered}$

$\begin{gathered}=\displaystyle \lim_{n\to \infty}(6+\frac{5}{2}\frac{(n+1)}{n}+\frac{1}{6}\frac{(n+1)(2n+1)}{n^2})\\\\=\displaystyle \lim_{n\to \infty}(6+\frac{5}{2}(1+\frac{1}{n})+\frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n}))\\\\\\=6+\dfrac{5}{2}+\dfrac{2}{6}=\dfrac{53}{6}\end{gathered}$

Hope it's helpful!!