Question

Calculate
$\huge{ \orange{ \displaystyle \lim_{n \to \infty} \frac{1}{n} \biggl \{ \displaystyle{} (m + 1) (m + 2). \: . \: . \: .(m + n) \biggl \} {}^{ \frac{1}{n} } }}$

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Answer 1

Solution−

Given expression is

$\rm :\longmapsto\:\displaystyle \lim_{n \to \infty} \frac{1}{n} \biggl \{ (m + 1)(m + 2). \: . \: . \: .(m + n) \bigg\} {}^{ \dfrac{1}{n} }$

Let assume that:—

$\rm :\longmapsto\:y = \displaystyle \lim_{n \to \infty} \frac{1}{n} \biggl \{ (m + 1)(m + 2) - - (m + n) \bigg\} {}^{ \dfrac{1}{n} }$

It can be rewritten as,

$\rm :\longmapsto\:y = \displaystyle \lim_{n \to \infty} \biggl \{ \frac{(m + 1)(m + 2) - - (m + n)}{ {n}^{n} } \bigg\} {}^{ \dfrac{1}{n} }$

$\rm :\longmapsto\:y = \displaystyle \lim_{n \to \infty} \biggl \{ \frac{(m + 1)(m + 2) - - (m + n)}{ {n}^{n} } \bigg\} {}^{ \dfrac{1}{n} }$

It can be further rewritten as,

$\rm :\longmapsto\:y = \displaystyle \lim_{n \to \infty} \biggl \{ \frac{(m + 1)(m + 2) - - (m + n)}{n.n. - - n } \bigg\} {}^{ \dfrac{1}{n} }$

$\rm :\longmapsto\:y = \displaystyle \lim_{n \to \infty} \biggl \{ \frac{m + 1}{n}. \frac{m + 2}{n} - - \frac{m + n}{n} \bigg\} {}^{ \dfrac{1}{n} }$

Taking log on both sides, we get

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty} log \biggl \{ \frac{m + 1}{n}. \frac{m + 2}{n} - - \frac{m + n}{n} \bigg\} {}^{ \dfrac{1}{n} }$

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} log \biggl \{ \frac{m + 1}{n}. \frac{m + 2}{n} - - \frac{m + n}{n} \bigg\}$

Can be rewritten as,

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ \frac{m + r}{n} \bigg\}$

Case - 1 When m = n

The above can be rewritten as,

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ \frac{n + r}{n} \bigg\}$

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ 1 + \frac{r}{n} \bigg\}$

Using Limit as a sum,

$\rm :\longmapsto\:logy = \displaystyle \int _{0}^{1} log \biggl \{ 1 + x \bigg\} \: dx$

On substituting, 1 + x = z, we get dx = dz

So,

$\rm :\longmapsto\:logy = \displaystyle \int _{1}^{2} log \biggl \{ z \bigg\} \: dz$

Using By parts, we get,

$\rm :\longmapsto\:logy = \displaystyle \bigg(zlogz - z \bigg) _{1}^{2}$

$\rm :\longmapsto\:logy = 2log2 - 1$

$\rm :\longmapsto\:logy = log4 - loge$

$\rm :\longmapsto\:logy = log\bigg |\dfrac{4}{e} \bigg|$

$\rm :\longmapsto\:y = \dfrac{4}{e}$

Case :- 2

When m > n

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ \frac{m + r}{n} \bigg\}$

Let we substitute m = n + h

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ \frac{n + h + r}{n} \bigg\}$

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ 1 + \frac{h}{n} + \frac{ r}{n} \bigg\}$

Limit does not exist.

Case - 3

When m < n

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ \frac{m + r}{n} \bigg\}$

Substituting m = n - h

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ \frac{n - h + r}{n} \bigg\}$

$\rm :\longmapsto\:logy = \displaystyle \lim_{n \to \infty}\dfrac{1}{n} \sum _{r = 1}^{n} log \biggl \{ 1 - \frac{h}{n} + \frac{ r}{n} \bigg\}$

Limit does not exist.