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Find the value of $\displaystyle \lim_{n \to \infty }\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+.....+\sqrt{n}}{{{n}^{\dfrac{3}{2}}}}$.

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Last updated date: 29th Feb 2024
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IVSAT 2024
Answer
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Hint: Divide both the numerator and the denominator with $\sqrt{n}$ and convert the given sum of infinite terms into a summation series of the form $\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{\dfrac{1}{n}f\left( \dfrac{r}{n} \right)}$. Now, in the next step convert this expression of limit into a definite integral by replacing $\left( \dfrac{r}{n} \right)$ with x and $\dfrac{1}{n}$ with $dx$. The upper and lower limits of the integral will be found by substituting r = n and r = 1 in the expression $\displaystyle \lim_{n \to \infty }\left( \dfrac{r}{n} \right)$ respectively and simplifying. Finally, use the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ and substitute the suitable limits to get the answer.

Complete step by step answer:
We have been provided with the expression $\displaystyle \lim_{n \to \infty }\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+.....+\sqrt{n}}{{{n}^{\dfrac{3}{2}}}}$ and we are asked to find its value. Here we need to convert the expression of limit into a definite integration to solve the question. Let us assume the given limit as I, so we have,
$\Rightarrow I=\displaystyle \lim_{n \to \infty }\dfrac{\sqrt{1}+\sqrt{2}+\sqrt{3}+.....+\sqrt{n}}{{{n}^{\dfrac{3}{2}}}}$
Dividing the numerator and the denominator with $\sqrt{n}$ we get,
\[\begin{align}
  & \Rightarrow I=\displaystyle \lim_{n \to \infty }\dfrac{\left( \dfrac{\sqrt{1}}{\sqrt{n}}+\dfrac{\sqrt{2}}{\sqrt{n}}+\dfrac{\sqrt{3}}{\sqrt{n}}+.....+\dfrac{\sqrt{n}}{\sqrt{n}} \right)}{\left( \dfrac{{{n}^{\dfrac{3}{2}}}}{\sqrt{n}} \right)} \\
 & \Rightarrow I=\displaystyle \lim_{n \to \infty }\dfrac{\left( \sqrt{\dfrac{1}{n}}+\sqrt{\dfrac{2}{n}}+\sqrt{\dfrac{3}{n}}+.....+\sqrt{\dfrac{n}{n}} \right)}{\left( \dfrac{{{n}^{\dfrac{3}{2}}}}{{{n}^{\dfrac{1}{2}}}} \right)} \\
\end{align}\]
Using the formula of exponents given as $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ in the denominator we get,
\[\begin{align}
  & \Rightarrow I=\displaystyle \lim_{n \to \infty }\dfrac{\left( \sqrt{\dfrac{1}{n}}+\sqrt{\dfrac{2}{n}}+\sqrt{\dfrac{3}{n}}+.....+\sqrt{\dfrac{n}{n}} \right)}{{{n}^{\dfrac{3}{2}-\dfrac{1}{2}}}} \\
 & \Rightarrow I=\displaystyle \lim_{n \to \infty }\dfrac{\left( \sqrt{\dfrac{1}{n}}+\sqrt{\dfrac{2}{n}}+\sqrt{\dfrac{3}{n}}+.....+\sqrt{\dfrac{n}{n}} \right)}{n} \\
 & \Rightarrow I=\displaystyle \lim_{n \to \infty }\dfrac{1}{n}\left( \sqrt{\dfrac{1}{n}}+\sqrt{\dfrac{2}{n}}+\sqrt{\dfrac{3}{n}}+.....+\sqrt{\dfrac{n}{n}} \right) \\
\end{align}\]
In the summation form we can write the expression as: -
\[\Rightarrow I=\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{\dfrac{1}{n}\left( \sqrt{\dfrac{r}{n}} \right)}\]
Now, we know that any limit of the form $\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{\dfrac{1}{n}f\left( \dfrac{r}{n} \right)}$ can be converted into a definite integral by replacing $\left( \dfrac{r}{n} \right)$ with x and $\dfrac{1}{n}$ with $dx$. So, the limit can be written as: -
\[\Rightarrow I=\int{\sqrt{x}dx}\]
Here, we need to determine the limits of this integral. To do this, we substitute r = n and r = 1 in the expression $\displaystyle \lim_{n \to \infty }\left( \dfrac{r}{n} \right)$ respectively and simplify. So the upper limit will be $\displaystyle \lim_{n \to \infty }\left( \dfrac{n}{n} \right)=1$ and the lower limit will be $\displaystyle \lim_{n \to \infty }\left( \dfrac{1}{n} \right)=0$. Therefore the integral becomes: -
$\Rightarrow I=\int_{0}^{1}{{{x}^{\dfrac{1}{2}}}dx}$
Using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ we get,
$\begin{align}
  & \Rightarrow I=\left[ \dfrac{{{x}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right]_{0}^{1} \\
 & \Rightarrow I=\left[ \dfrac{{{x}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]_{0}^{1} \\
 & \Rightarrow I=\dfrac{2}{3}\left[ {{x}^{\dfrac{3}{2}}} \right]_{0}^{1} \\
\end{align}$
Substituting the suitable limits and simplifying we get,
$\begin{align}
  & \Rightarrow I=\dfrac{2}{3}\left[ {{1}^{\dfrac{3}{2}}}-{{0}^{\dfrac{3}{2}}} \right] \\
 & \Rightarrow I=\dfrac{2}{3}\left[ 1 \right] \\
 & \therefore I=\dfrac{2}{3} \\
\end{align}$
Hence, the value of the given limit is $\dfrac{2}{3}$.

Note: It is necessary to convert the given limit into a definite integral because there is no direct formula in limits which can be applied to solve the question. Do not apply L'Hospital's rule to get the answer. Remember that the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$ cannot be applied for n = -1 because if n = -1 then the function becomes $\dfrac{1}{x}$ whose integral is $\ln x$.