Question

The value of \[ \lim_{n \to \infty} \left\{ \frac{1}{n+m} + \frac{1}{n+2m} + \frac{1}{n+3m} + \dots + \frac{1}{n+nm} \right\} \] is:

• A. $\dfrac{\log_e(m)}{m}$
• B. $\dfrac{\log_e(1+m)}{1+m}$
• C. $\dfrac{\log_e(1+m)}{m}$
• D. $\dfrac{\log_e(1+m)}{1-m}$

Hint:

Whenever a limit contains a summation with terms involving: \[ \frac{r}{n} \] try converting it into a Riemann integral using: \[ \frac{1}{n}\sum f\left(\frac{r}{n}\right) \rightarrow \int_0^1 f(x)\,dx \] Factoring out $n$ from the denominator is usually the key first step.

Answer 1

The Correct Option is C

Concept: Limits involving large summations are often evaluated using the concept of Riemann sums. The standard result is: \[ \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) = \int_0^1 f(x)\,dx \] Thus, the main objective is to rewrite the given expression in the form: \[ \frac{1}{n}\sum f\left(\frac{r}{n}\right) \] so that it can be converted into a definite integral.

Step 1: Writing the expression in sigma notation.
The given summation is: \[ S = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n+rm} \] Now factor out $n$ from the denominator: \[ S = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n\left(1+\frac{rm}{n}\right)} \] \[ S = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} \frac{1}{1+m\left(\frac{r}{n}\right)} \] Now the expression is in standard Riemann sum form.

Step 2: Converting the summation into a definite integral.
Compare with: \[ \frac{1}{n}\sum_{r=1}^{n} f\left(\frac{r}{n}\right) \] Here, \[ f(x)=\frac{1}{1+mx} \] Therefore, \[ S= \int_0^1 \frac{1}{1+mx}\,dx \]

Step 3: Evaluating the integral.
We evaluate: \[ I= \int_0^1 \frac{1}{1+mx}\,dx \] Using the standard integral: \[ \int \frac{1}{a+bx}\,dx = \frac{1}{b}\log_e(a+bx) \] we get: \[ I= \left[ \frac{1}{m}\log_e(1+mx) \right]_0^1 \] Substituting the limits: \[ I= \frac{1}{m} \left[ \log_e(1+m)-\log_e(1) \right] \] Since, \[ \log_e(1)=0 \] therefore, \[ I= \frac{\log_e(1+m)}{m} \]

Step 4: Final conclusion.
Hence, \[ \boxed{ \lim_{n \to \infty} \left( \frac{1}{n+m} + \frac{1}{n+2m} + \dots + \frac{1}{n+nm} \right) = \frac{\log_e(1+m)}{m} } \] Therefore, the correct answer is: \[ \boxed{(C)\ \dfrac{\log_e(1+m)}{m}} \]