Question

If we have a summation series $I=\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}{\displaystyle \frac{k+1}{x(x+1)}}dx$, then which of the following options is correct.
(This question has multiple correct options)
(a) $I<{\displaystyle \frac{49}{50}}$
(b) $I>{\log }_{e}99$
(c) $I>{\displaystyle \frac{49}{50}}$
(d) $I<{\log }_{e}99$

Answer 1

Hint: In any definite integral, the value of the variable of the integral lies between the limits of the integral. This is a very important property of integrals and we will make use of this property to solve the given question. In our question, the variable of integration is x and the limits are k and k + 1. We will write them in the form inequalities and manipulate the inequality to get the range of the expression of inequality. Once we get this range, we will apply summation and integration throughout and solve the integrals at the extreme values of the expression. Then we will verify which of the given options are correct with regards to our solution.

Complete step-by-step solution:
The integral given to us is as follows:
$I=\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}{\displaystyle \frac{k+1}{x(x+1)}}dx$
The upper limit of the integral is k + 1 and the lower limit is k. The variable of integration is x.
Therefore, we can safely say that $kWe will take the later inequality which is $xWe will divide both sides with x.
$\Rightarrow 1<{\displaystyle \frac{k+1}{x}}......\left(1\right)$
Now we will take the former inequality, which is $kWe shall add 1 to both sides of the inequality.
$\Rightarrow k+1It is to be noted that the value of x is always positive because x is greater than or equal to the lower limit k, whose minimum value is 1.
Now, we will divide both sides with (x + 1).
$\Rightarrow {\displaystyle \frac{k+1}{x+1}}<1......\left(2\right)$
Now, we will divide both sides of (1) by (x + 1) and both sides of (2) by x.
$\Rightarrow {\displaystyle \frac{1}{x+1}}<{\displaystyle \frac{k+1}{x(x+1)}}......\left(3\right)$
$\Rightarrow {\displaystyle \frac{k+1}{x(x+1)}}<{\displaystyle \frac{1}{x}}......\left(4\right)$
Now, form (3) and (4), we can form the following inequality.
$\Rightarrow {\displaystyle \frac{1}{x+1}}<{\displaystyle \frac{k+1}{x(x+1)}}<{\displaystyle \frac{1}{x}}$
Now, we will apply integral with upper limit k + 1 and lower limit k throughout the inequality.
$$\Rightarrow \underset{k}{\overset{k+1}{\int }}{\displaystyle \frac{1}{x+1}}dx<\underset{k}{\overset{k+1}{\int }}{\displaystyle \frac{k+1}{x(x+1)}}dx<\underset{k}{\overset{k+1}{\int }}{\displaystyle \frac{1}{x}}dx$$
Now, we will solve the integrals of extreme values and keep the middle term as it is.
$$\Rightarrow {\log }_e\left({\displaystyle \frac{k+2}{k+1}}\right)<\underset{k}{\overset{k+1}{\int }}{\displaystyle \frac{k+1}{x(x+1)}}dx<{\log }_e\left({\displaystyle \frac{k+1}{k}}\right)$$
Now, we will apply summation from k=1 to 98 throughout the inequality.
$$\Rightarrow \sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+2}{k+1}}\right)<\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}{\displaystyle \frac{k+1}{x(x+1)}}dx<\sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+1}{k}}\right)$$
But we know that $I=\sum _{k=1}^{98}\underset{k}{\overset{k+1}{\int }}{\displaystyle \frac{k+1}{x(x+1)}}dx$
$$\Rightarrow \sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+2}{k+1}}\right)<I<\sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+1}{k}}\right)$$
Now, we will consider the left most expression.
$$\begin{array}{rl}& \Rightarrow \sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+2}{k+1}}\right)={\log }_e\left(3\right)-{\log }_e\left(2\right)+{\log }_e\left(4\right)-{\log }_e\left(3\right)+...+lo{g}_e\left(100\right)-{\log }_e\left(99\right)\\ & \Rightarrow \sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+2}{k+1}}\right)={\log }_e\left(100\right)-{\log }_e\left(2\right)\\ & \Rightarrow \sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+2}{k+1}}\right)={\log }_e\left(50\right)\end{array}$$
Now, we shall consider the right most expression.
$$\begin{array}{rl}& \Rightarrow \sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+1}{k}}\right)={\log }_e\left(2\right)-{\log }_e\left(1\right)+{\log }_e\left(3\right)-{\log }_e\left(2\right)+...+{\log }_e\left(99\right)-{\log }_e\left(98\right)\\ & \Rightarrow \sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+1}{k}}\right)={\log }_e\left(99\right)-{\log }_e\left(1\right)\\ & \Rightarrow \sum _{k=1}^{98}{\log }_e\left({\displaystyle \frac{k+1}{k}}\right)={\log }_e\left(99\right)\end{array}$$
Therefore, the inequality modifies as follows:
$$\Rightarrow {\log }_e\left(50\right)<I<{\log }_e\left(99\right)$$
Hence, it is clear that option (d) is correct.
Now value of ${\log }_{e}50=3.912$
Therefore, $I>3.912$
Thus, $I>{\displaystyle \frac{49}{50}}$
Hence, the option (c) is also correct.

Note: It is advised to be aware of the options. From the options, it is clear that all we need to find is inequality and not the actual value of the integration. Thus, we do need to find the actual value of the integral, which in fact is very complex to solve.