Question

$\displaystyle \sf \red{\lim_{n \to \infty} \frac{1}{n} \biggl \{ \displaystyle{\sf \prod_{r = 1} ^{n}} \sf (m + r) \biggl \} {}^{ \frac{1}{n} } }$​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle \sf {\lim_{n \to \infty} \frac{1}{n} \biggl \{ \displaystyle{\sf \prod_{r = 1} ^{n}} \sf (m + r) \biggl \} {}^{ \frac{1}{n} } }$

Let assume that

$\rm :\longmapsto\:y = \displaystyle \sf {\lim_{n \to \infty} \frac{1}{n} \biggl \{ \displaystyle{\sf \prod_{r = 1} ^{n}} \sf (m + r) \biggl \} {}^{ \frac{1}{n} } }$

can be rewritten as

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

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

can be further rewritten as

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

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

Now, taking log on both sides, we get

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

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

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

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

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

which can be rewritten as

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

Now, We consider 3 different cases.

Case :- 1 When m = n

The above expression can be rewritten as

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

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

Now, using Limit as a sum from definite integrals, we have

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

Now, to evaluate this integral, we use Method of Substitution.

So, Substitute

$\rm :\longmapsto\:1 + x = z$

$\rm :\longmapsto\:dx = dz$

So, above integral can be rewritten as

$\rm :\longmapsto\:logy = \displaystyle\int_{1}^2logz \: dz$

Now, using integration by parts, we get

$\rm :\longmapsto\:logy = \bigg|z \: logz - z\bigg| _{1}^2$

$\rm :\longmapsto\:logy = 2 \: log2 - 2 - (0 - 1)$

$\rm :\longmapsto\:logy = \: log {2}^{2} - 1$

$\rm :\longmapsto\:logy = \: log 4 - loge$

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

$\bf\implies \:y = \dfrac{4}{e}$

Hence,

$\red{\bf\implies \:\boxed{\tt{ \:\displaystyle \sf {\lim_{n \to \infty} \frac{1}{n} \biggl \{ \displaystyle{\sf \prod_{r = 1} ^{n}} \sf (m + r) \biggl \} {}^{ \dfrac{1}{n} } } \: = \: \frac{4}{e} \: \: }}}$

Case :- 2 When m < n

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

On substituting m = n - h, we get

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

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

It implies, Limit doesnot exist.

Case :- 3 When m > n

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

Substitute m = n + h, we get

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

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

It implies, Limit doesnot exist.

Answer 2

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

Given expression is

$\rm :\longmapsto\:\displaystyle \sf {\lim_{n \to \infty} \frac{1}{n} \biggl \{ \displaystyle{\sf \prod_{r = 1} ^{n}} \sf (m + r) \biggl \} {}^{ \frac{1}{n} } }$

Let assume that

$\rm :\longmapsto\:y = \displaystyle \sf {\lim_{n \to \infty} \frac{1}{n} \biggl \{ \displaystyle{\sf \prod_{r = 1} ^{n}} \sf (m + r) \biggl \} {}^{ \frac{1}{n} } }$

can be rewritten as

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

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

can be further rewritten as

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

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

Now, taking log on both sides, we get

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

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

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

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

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

which can be rewritten as

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

Now, We consider 3 different cases.

Case :- 1 When m = n

The above expression can be rewritten as

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

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

Now, using Limit as a sum from definite integrals, we have

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

Now, to evaluate this integral, we use Method of Substitution.

So, Substitute

$\rm :\longmapsto\:1 + x = z$

$\rm :\longmapsto\:dx = dz$

So, above integral can be rewritten as

$\rm :\longmapsto\:logy = \displaystyle\int_{1}^2logz \: dz$

Now, using integration by parts, we get

$\rm :\longmapsto\:logy = \bigg|z \: logz - z\bigg| _{1}^2$

$\rm :\longmapsto\:logy = 2 \: log2 - 2 - (0 - 1)$

$\rm :\longmapsto\:logy = \: log {2}^{2} - 1$

$\rm :\longmapsto\:logy = \: log 4 - loge$

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

$\bf\implies \:y = \dfrac{4}{e}$

Hence,

$\purple{\bf\implies \:\boxed{\tt{ \:\displaystyle \sf {\lim_{n \to \infty} \frac{1}{n} \biggl \{ \displaystyle{\sf \prod_{r = 1} ^{n}} \sf (m + r) \biggl \} {}^{ \dfrac{1}{n} } } \: = \: \frac{4}{e} \: \: }}}$

Case :- 2 When m < n

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

On substituting m = n - h, we get

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

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

It implies, Limit doesnot exist.