Question

By the definition of the definite integral, the value of $\displaystyle \lim_{n \to \infty }\left( \dfrac{{{1}^{4}}}{{{1}^{5}}+{{n}^{5}}}+\dfrac{{{2}^{4}}}{{{2}^{5}}+{{n}^{5}}}+\dfrac{{{3}^{4}}}{{{3}^{5}}+{{n}^{5}}}+......+\dfrac{{{n}^{4}}}{{{n}^{5}}+{{n}^{5}}} \right)$ is
(a) log 2
(b) $\dfrac{1}{5}\log 2$
(c) $\dfrac{1}{4}\log 2$
(d) $\dfrac{1}{3}\log 2$

Answer 1

Hint: Divide all the numerators and denominators of different terms by ${{n}^{5}}$ . In the numerator write the expressions as $\dfrac{1}{n}\left( \dfrac{{{1}^{4}}}{{{n}^{4}}} \right),\dfrac{1}{n}\left( \dfrac{{{2}^{4}}}{{{n}^{4}}} \right),\dfrac{1}{n}\left( \dfrac{{{3}^{4}}}{{{n}^{4}}} \right)$ and so on. Now, write the expression using sigma sign $\Sigma $ as $\dfrac{1}{n}f\left( \dfrac{r}{n} \right)$ , where r ranges from 1 to n. Now, convert the expression into a definite integral by replacing $\dfrac{r}{n}$ with x and $\dfrac{1}{n}$ with dx. Determine the limits of the integral by substituting r = 0 and r = n for $n \to \infty $ in the ratio $\dfrac{r}{n}$. Solve the integral to get the answer.

Complete step-by-step solution:
Here, we have been asked to find the value of the expression $\displaystyle \lim_{n \to \infty }\left( \dfrac{{{1}^{4}}}{{{1}^{5}}+{{n}^{5}}}+\dfrac{{{2}^{4}}}{{{2}^{5}}+{{n}^{5}}}+\dfrac{{{3}^{4}}}{{{3}^{5}}+{{n}^{5}}}+......+\dfrac{{{n}^{4}}}{{{n}^{5}}+{{n}^{5}}} \right)$. So, let us assume its value to be $I$. so, we have
$\Rightarrow I=\displaystyle \lim_{n \to \infty }\left( \dfrac{{{1}^{4}}}{{{1}^{5}}+{{n}^{5}}}+\dfrac{{{2}^{4}}}{{{2}^{5}}+{{n}^{5}}}+\dfrac{{{3}^{4}}}{{{3}^{5}}+{{n}^{5}}}+......+\dfrac{{{n}^{4}}}{{{n}^{5}}+{{n}^{5}}} \right)$
Dividing the numerators and denominators of all the terms with ${{n}^{5}}$, we get,
$\Rightarrow I=\displaystyle \lim_{n \to \infty }\left( \dfrac{\left( \dfrac{{{1}^{4}}}{{{n}^{5}}} \right)}{\dfrac{{{1}^{5}}}{{{n}^{5}}}+1}+\dfrac{\left( \dfrac{{{2}^{4}}}{{{n}^{5}}} \right)}{\dfrac{{{2}^{5}}}{{{n}^{5}}}+1}+\dfrac{\left( \dfrac{{{3}^{4}}}{{{n}^{5}}} \right)}{\dfrac{{{3}^{5}}}{{{n}^{5}}}+1}+......+\dfrac{\left( \dfrac{{{n}^{4}}}{{{n}^{5}}} \right)}{\dfrac{{{n}^{5}}}{{{n}^{5}}}+1} \right)$
The above expression can be written as
\[\begin{align}
  & \Rightarrow I=\displaystyle \lim_{n \to \infty }\left( \dfrac{\dfrac{1}{n}{{\left( \dfrac{1}{n} \right)}^{4}}}{{{\left( \dfrac{1}{n} \right)}^{5}}+1}+\dfrac{\dfrac{1}{n}{{\left( \dfrac{2}{n} \right)}^{4}}}{{{\left( \dfrac{2}{n} \right)}^{5}}+1}+\dfrac{\dfrac{1}{n}{{\left( \dfrac{3}{n} \right)}^{4}}}{{{\left( \dfrac{3}{n} \right)}^{5}}+1}+......+\dfrac{\dfrac{1}{n}{{\left( \dfrac{n}{n} \right)}^{4}}}{{{\left( \dfrac{n}{n} \right)}^{5}}+1} \right) \\
 & \Rightarrow I=\displaystyle \lim_{n \to \infty }\left[ \dfrac{1}{n}\left( \dfrac{{{\left( \dfrac{1}{n} \right)}^{4}}}{{{\left( \dfrac{1}{n} \right)}^{5}}+1}+\dfrac{{{\left( \dfrac{2}{n} \right)}^{4}}}{{{\left( \dfrac{2}{n} \right)}^{5}}+1}+\dfrac{{{\left( \dfrac{3}{n} \right)}^{4}}}{{{\left( \dfrac{3}{n} \right)}^{5}}+1}+......+\dfrac{{{\left( \dfrac{n}{n} \right)}^{4}}}{{{\left( \dfrac{n}{n} \right)}^{5}}+1} \right) \right] \\
\end{align}\]
Using the summation notation $\Sigma $, we can write,
\[\Rightarrow I=\displaystyle \lim_{n \to \infty }\sum\limits_{r=1}^{n}{\dfrac{1}{n}\times \left[ \dfrac{{{\left( \dfrac{r}{n} \right)}^{4}}}{{{\left( \dfrac{r}{n} \right)}^{5}}+1} \right]}\]
The R.H.S of the above expression is of the form $\dfrac{1}{n}f\left( \dfrac{r}{n} \right)$, where $f\left( \dfrac{r}{n} \right)$ is a function of $\left( \dfrac{r}{n} \right)$. To convert this expression into fractional form, we replace $\left( \dfrac{r}{n} \right)$ with x and with $\dfrac{1}{n}$ dx. So, we have
$\Rightarrow I=\int\limits_{a}^{b}{\left( \dfrac{{{x}^{4}}}{{{x}^{5}}+1} \right)}dx$
Here, a and b are the lower and upper limits of the integral. Let us determine the values of a and b. Since n is tending to infinity, so we have
(i) At r = 0,
$\begin{align}
  & \Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{r}{n}=\dfrac{0}{\infty }=0 \\
 & \Rightarrow a=0 \\
\end{align}$
(ii) At r = n,
$\begin{align}
  & \Rightarrow \displaystyle \lim_{n \to \infty }\dfrac{r}{n}=\displaystyle \lim_{n \to \infty }\dfrac{n}{n}=1 \\
 & \Rightarrow b=1 \\
\end{align}$
So the expression for I becomes,
$\Rightarrow I=\int\limits_{0}^{1}{\left( \dfrac{{{x}^{4}}}{{{x}^{5}}+1} \right)}dx$
Let us assume ${{x}^{5}}+1=k$ . Therefore, on differentiating both sides, we get
$\begin{align}
  & \Rightarrow 5{{x}^{4}}dx=dk \\
 & \Rightarrow {{x}^{4}}dx=\dfrac{dk}{5} \\
\end{align}$
Substituting the above relation in I, we get
$\begin{align}
  & \Rightarrow I=\int\limits_{0}^{1}{\dfrac{dk}{5k}} \\
 & \Rightarrow I=\dfrac{1}{5}\int\limits_{0}^{1}{\dfrac{dk}{k}} \\
\end{align}$
Now we know that $\int{\dfrac{dx}{x}=\ln x}$ , so we have
$\Rightarrow I=\dfrac{1}{5}\left[ \ln k \right]_{0}^{1}$
Substituting the value of ‘k’ back, we get
$\Rightarrow I=\dfrac{1}{5}\left[ \ln \left( {{x}^{5}}+1 \right) \right]_{0}^{1}$
Now, substituting the limits, we get
$\begin{align}
  & \Rightarrow I=\dfrac{1}{5}\left[ \ln \left( {{1}^{5}}+1 \right)-\ln \left( {{0}^{5}}+1 \right) \right] \\
 & \Rightarrow I=\dfrac{1}{5}\left[ \ln 2-\ln 1 \right] \\
 & \because \ln 1=0 \\
 & \Rightarrow I=\dfrac{1}{5}\left[ \ln 2-0 \right] \\
 & \therefore I=\dfrac{1}{5}\ln 2 \\
\end{align}$
Hence, option (b) is the correct answer.

Note: One may note that when we have substituted ${{x}^{5}}+1=k$, we have not changed the limits of the integration. This is because at last, we have substituted back the value of k in terms of x. If we would not have substituted back the value of k then only the change of limits would have been mandatory. You must remember the process to change the limit into a definite integral otherwise you will not be able to solve this question.