Question

Let ${{S}_{n}}=\dfrac{1}{1+\sqrt{n}}+\dfrac{1}{2+\sqrt{2n}}+\ldots +\dfrac{1}{n+\sqrt{{{n}^{2}}}}$, then find the limit of $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}$.

Answer 1

Hint: Convert summation to integral using Riemann integral concept and then apply the limits.

Consider the given expression,
${{S}_{n}}=\dfrac{1}{1+\sqrt{n}}+\dfrac{1}{2+\sqrt{2n}}+\ldots +\dfrac{1}{n+\sqrt{{{n}^{2}}}}$
This can be converted to summation as,
{{S}_{n}}=\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{1}{r+\sqrt{rn}}
Now we will apply limits, we get
\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{1}{r+\sqrt{rn}}
Now dividing numerator and denominator by n, we get
\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\dfrac{\left( r+\sqrt{rn} \right)}{n}}
In the denominator we will separate the terms, then we get
\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\dfrac{r}{n}+\dfrac{\sqrt{rn}}{n}}
Taking the $'n'$ inside the root we get,
\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\dfrac{r}{n}+\sqrt{\dfrac{rn}{{{n}^{2}}}}}
\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\dfrac{r}{n}+\sqrt{\dfrac{r}{n}}}\ldots \ldots \left( i \right)
Now we know the way to solve the summation limit is to convert the summation into integral.
For this first let us assume
$\dfrac{r}{n}=x\Rightarrow \dfrac{1}{n}=dx$
Let’s find the limits,
When $r=1$, then $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{r}{n}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}=0$ , therefore, $x=0$.
When $r=n$ , then $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{r}{n}=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{n}{n}=1$, therefore, $x=1$.
Considering these values the summation can be written as integral form. We get,
\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\dfrac{r}{n}+\sqrt{\dfrac{r}{n}}}=\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}\ldots \ldots \ldots \left( ii \right)
Now we will find the integration as follows:
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{\sqrt{x}\left( \sqrt{x}+1 \right)}\ldots \ldots .\left( iii \right)
Let’s substitute,
$u=\sqrt{x}+1$
Differentiating, we get
$du=\dfrac{1}{2\sqrt{x}}dx\Rightarrow dx=2\sqrt{x}du$
Substituting these values in equation (iii), we get
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{2\sqrt{x}du}{\sqrt{x}\left( u \right)}
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=2\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{1}{u}du
But we know, $\mathop{\int }^{}\dfrac{1}{u}=\ln u$ , so above equation becomes,
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=2\left[ lnu \right]_{0}^{1}
Substituting back the value of u, we get
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=2\left[ \ln (\sqrt{x}+1) \right]_{0}^{1}
Applying the upper and lower bounds, we get
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=2\left[ \ln (\sqrt{1}+1 \right)\left] -2 \right[\ln (\sqrt{0}+1)]
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=2\left[ \ln 2\left] -2 \right[\ln (1 \right)]
But we know, $ln1=0$, so we get
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=2\left[ ln2 \right]-0
\underset{0}{\overset{1}{\mathop \int }}\,\dfrac{dx}{x+\sqrt{x}}=\left[ \ln {{2}^{2}} \right]=\ln 4
Substituting this in equation (ii), we get
\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\dfrac{r}{n}+\sqrt{\dfrac{r}{n}}}=\ln 4
Substituting this value in equation (i), we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\ln 4$
As the RHS is free of variable, so we can remove the limit, so we get
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\ln 4$

Note: The possibility for the mistake is that in the following equation,
\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\dfrac{\dfrac{1}{n}}{\dfrac{r}{n}+\sqrt{\dfrac{rn}{{{n}^{2}}}}}
we see that the n2 term becomes ‘n’ when taken out of the square root.
So, instead of dividing by n2 we can just divide the ‘r’ by ‘n’ and we won’t get the same answer.
One more possible mistake is when converting the summation to integral, the student will get confused on how to get the limit of integral.