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:
Show 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.
Step 1: Understanding the Concept:
Limits of summations as \( n \to \infty \) can frequently be computed by recognizing the sum as a Riemann sum of a definite integral.
The definition of a definite integral is:
\[ \int_0^1 f(x) dx = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^{n} f\left(\frac{r}{n}\right) \]
The key is to rewrite the given sum in a way that matches this template.
We look for a factor of \( 1/n \) and terms involving the ratio \( r/n \). Step 2: Key Formula or Approach:
1. Summation notation: \( S = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n + rm} \).
2. Factoring out \( n \): \( S = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{1 + m(r/n)} \). Step 3: Detailed Explanation:
Let's express the limit more formally:
\[ L = \lim_{n \to \infty} \left[ \frac{1}{n+m} + \frac{1}{n+2m} + \dots + \frac{1}{n+nm} \right] \]
Rewrite the terms using a summation index \( r \):
\[ L = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n + rm} \]
To convert this to the standard Riemann form, factor \( n \) out from the denominator:
\[ L = \lim_{n \to \infty} \sum_{r=1}^n \frac{1}{n (1 + m \cdot \frac{r}{n})} \]
Move the constant factor \( 1/n \) outside the summation sign:
\[ L = \lim_{n \to \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{1 + m(\frac{r}{n})} \]
This is exactly in the form \( \frac{1}{n} \sum f(r/n) \) where \( f(x) = \frac{1}{1 + mx} \).
The limits of integration go from \( r/n = 1/n \approx 0 \) to \( r/n = n/n = 1 \).
\[ L = \int_0^1 \frac{1}{1 + mx} dx \]
To solve this integral, perform a basic substitution or use the standard logarithmic form:
\[ L = \left[ \frac{\log_e(1 + mx)}{m} \right]_0^1 \]
Evaluate at the limits:
\[ L = \frac{1}{m} [\log_e(1 + m \cdot 1) - \log_e(1 + m \cdot 0)] \]
\[ L = \frac{1}{m} [\log_e(1 + m) - \log_e(1)] \]
Since \( \log_e(1) = 0 \):
\[ L = \frac{\log_e(1 + m)}{m} \] Step 4: Final Answer:
The value of the limit is \( \frac{\log_e(1+m)}{m} \).
This is Option (C).