Question

The limit, when n tends to infinity, of the series $\dfrac{\sqrt{n}}{\sqrt{{{n}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4 \right)}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+8 \right)}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+12 \right)}^{3}}}}+..................\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4n \right)}^{3}}}}$ is:

Answer 1

Hint: The question is related to a concept of limit of a sum. If you observe you will find that the series can be represented as $\sum\limits_{r=0}^{n}{\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4r \right)}^{3}}}}}$ and represent it in the form \[\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{a}^{b}{f\left( \dfrac{r}{n} \right)}=\int\limits_{\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{a}{n}}^{\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{b}{n}}{f(x)dx}\] . Now use the simple formulas related to integration to get the answer to the above question.

Complete step-by-step answer:
The series given in the question is:
$\dfrac{\sqrt{n}}{\sqrt{{{n}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4 \right)}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+8 \right)}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+12 \right)}^{3}}}}+..................\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4n \right)}^{3}}}}$
If we observe carefully, we will find that the numerator of the terms in the series are constant while the denominators follow a pattern. So, if we represent this in sigma form, i.e., more systematic form, we get
 $\dfrac{\sqrt{n}}{\sqrt{{{n}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4 \right)}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+8 \right)}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+12 \right)}^{3}}}}+..................\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4n \right)}^{3}}}}=\sum\limits_{r=0}^{n}{\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4r \right)}^{3}}}}}$
Now as it is given that n tends to infinity, so we will take limit on both sides of the equation. On doing so, we get
\[\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{\sqrt{n}}{\sqrt{{{n}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4 \right)}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+8 \right)}^{3}}}}+\dfrac{\sqrt{n}}{\sqrt{{{\left( n+12 \right)}^{3}}}}+..................\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4n \right)}^{3}}}} \right)=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=0}^{n}{\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4r \right)}^{3}}}}}\]
Now we will solve the right-hand side of the above equation.
\[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=0}^{n}{\dfrac{\sqrt{n}}{\sqrt{{{\left( n+4r \right)}^{3}}}}}\]
\[=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=0}^{n}{\dfrac{\sqrt{n}}{\sqrt{{{n}^{3}}{{\left( 1+4\dfrac{r}{n} \right)}^{3}}}}}\]
\[=\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=0}^{n}{\dfrac{1}{n}\times \dfrac{1}{\sqrt{{{\left( 1+4\dfrac{r}{n} \right)}^{3}}}}}\]
As $\dfrac{1}{n}$ is costant, we can take $\dfrac{1}{n}$ outside sigma. On doing so, we get
\[=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=0}^{n}{\dfrac{1}{\sqrt{{{\left( 1+4\dfrac{r}{n} \right)}^{3}}}}}\]
Now, according to the concept of limit of a sum, we know that \[\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{a}^{b}{f\left( \dfrac{r}{n} \right)}=\int\limits_{\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{a}{n}}^{\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{b}{n}}{f(x)dx}\] . So, our expression becomes:
\[\int\limits_{0}^{1}{\dfrac{1}{\sqrt{{{\left( 1+4x \right)}^{3}}}}}dx\]
\[=\int\limits_{0}^{1}{\dfrac{1}{{{\left( 1+4x \right)}^{\dfrac{3}{2}}}}}dx\]
Now we know that the integral of $\dfrac{1}{{{\left( ax+b \right)}^{c}}}=\dfrac{-1}{a\left( c-1 \right){{\left( ax+b \right)}^{c-1}}}$ . Using this for our integral, we get
\[\left. \dfrac{-1}{4\times \dfrac{1}{2}{{\left( 1+4x \right)}^{\dfrac{1}{2}}}} \right|_{0}^{1}\]
\[=-\left( \dfrac{1}{2{{\left( 1+4 \right)}^{\dfrac{1}{2}}}}-\dfrac{1}{2{{\left( 1 \right)}^{\dfrac{1}{2}}}} \right)\]
\[=-\left( \dfrac{1}{2\sqrt{5}}-\dfrac{1}{2} \right)=\dfrac{1}{2}-\dfrac{1}{2\sqrt{5}}\]
Therefore, the answer to the above question is \[\dfrac{1}{2}-\dfrac{1}{2\sqrt{5}}\] .

Note: In questions related to the concept of limit of a sum, the most important thing is to find the general term, as mostly the general term is based on observation and not on any mathematical equations or standards. Also, while applying the concept of limit of a sum, there should not be any r related term appearing independently, all the r related terms should have n as their denominators. If you are not comfortable with the formula of integration used, you can integrate by substituting 4x+1=u and solving the integral.