Question

Find the value of $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}$ when
     ${{S}_{n}}=\frac{1}{1+\sqrt{n}}+\frac{1}{2+\sqrt{2n}}+\ldots +\frac{1}{n+\sqrt{{{n}^{2}}}}$ .

Answer 1

Consider the given expression,
     ${{S}_{n}}=\frac{1}{1+\sqrt{n}}+\frac{1}{2+\sqrt{2n}}+\ldots +\frac{1}{n+\sqrt{{{n}^{2}}}}$
This can be converted to summation as,
     \[{{S}_{n}}=\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{1}{r+\sqrt{rn}}\]
 Now we will apply limits, we get
     $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{1}{r+\sqrt{rn}}$
Dividing numerator and denominator by n, we get
     $\underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{\frac{1}{n}}{\frac{\left( r+\sqrt{rn} \right)}{n}}$
     $\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{\frac{1}{n}}{\frac{r}{n}+\frac{\sqrt{rn}}{n}}$
     $\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{\frac{1}{n}}{\frac{r}{n}+\sqrt{\frac{rn}{{{n}^{2}}}}}$
     $\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{\frac{1}{n}}{\frac{r}{n}+\sqrt{\frac{r}{n}}}\ldots \ldots \left( i \right)$ \[\]
Now let,
     $\frac{r}{n}=x\Rightarrow \frac{1}{n}=dx$
Let’s find the limits,
When \[r=1\Rightarrow x=0\]
When $r=n\Rightarrow x=1$
Considering these values the summation can be written as integral form. We get,
     $\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{\frac{1}{n}}{\frac{r}{n}+\sqrt{\frac{r}{n}}}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}\ldots \ldots \ldots \left( ii \right)$
Now we will find the integration as follows:
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{\sqrt{x}\left( \sqrt{x}+1 \right)}\ldots \ldots .\left( iii \right)$
Let’s substitute,
     $u=\sqrt{x}+1$
Differentiating, we get
     $du=\frac{1}{2\sqrt{x}}dx\Rightarrow dx=2\sqrt{x}du$
Substituting these values in equation (iii), we get
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=\underset{0}{\overset{1}{\mathop \int }}\,\frac{2\sqrt{x}du}{\sqrt{x}\left( u \right)}$
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=2\underset{0}{\overset{1}{\mathop \int }}\,\frac{1}{u}du$
But we know, $\mathop{\int }^{}\frac{1}{u}=\ln u$ , so above equation becomes,
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=2\left[ lnu \right]_{0}^{1}$
Substituting back the value of u, we get
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=2\left[ \ln (\sqrt{x}+1) \right]_{0}^{1}$
Applying the upper and lower bounds, we get
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=2\left[ \ln (\sqrt{1}+1 \right)\left] -2 \right[\ln (\sqrt{0}+1)]$
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=2\left[ \ln 2\left] -2 \right[\ln (1 \right)]$
But we know, $ln1=0$, so we get
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=2\left[ ln2 \right]-0$
     $\underset{0}{\overset{1}{\mathop \int }}\,\frac{dx}{x+\sqrt{x}}=\left[ \ln {{2}^{2}} \right]=\ln 4$
Substituting this in equation (ii), we get
     $\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{\frac{1}{n}}{\frac{r}{n}+\sqrt{\frac{r}{n}}}=\ln 4$
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,
     $\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,{{S}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\underset{r=1}{\overset{n}{\mathop \sum }}\,\frac{\frac{1}{n}}{\frac{r}{n}+\sqrt{\frac{rn}{{{n}^{2}}}}}$
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.