Question

If \( k \in N \) then \( \lim_{n\to\infty} \left[ \frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \dots + \frac{1}{kn} \right] = \) (Note: The last term should be \( \frac{1}{n+ (k-1)n} = \frac{1}{kn} \) or sum up to \(n+(k-1)n\). The given form \(1/kn\) as the endpoint of the sum means sum from \(r=1\) to \((k-1)n\). The sum is usually \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). If the last term is \( \frac{1}{kn} \), it means \( n+r = kn \implies r = (k-1)n \). So it's \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \).) Let's assume the sum goes up to \( \frac{1}{n+(k-1)n} = \frac{1}{kn} \). So the sum is \( \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \). No, this seems to be \( \frac{1}{n+1} + \dots + \frac{1}{n+(kn-n)} \). The sum should be written as \( \sum_{i=1}^{(k-1)n} \frac{1}{n+i} \). The dots imply the denominator goes up. The last term is \( \frac{1}{kn} \). This means the sum is actually \( \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{n+(k-1)n} \). The number of terms is \( (k-1)n \).

• A. \( \log(k+1) \)
• B. \( \log k \)
• C. \( \log(k+5) \)
• D. \( \log(k+1) - \log 6 \)

Hint:

Limits of sums can often be converted to definite integrals using the Riemann sum definition: \( \lim_{n\to\infty} \frac{1}{n} \sum_{r=an}^{bn} f\left(\frac{r}{n}\right) = \int_{a}^{b} f(x) dx \) (if \(a,b\) are constants or based on limits of \(r/n\)). In this case, rewrite the sum as \( \lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{(k-1)n} \frac{1}{1+r/n} \). The limits of integration for \(x=r/n\) are from \( \lim_{n\to\infty} 1/n = 0 \) to \( \lim_{n\to\infty} (k-1)n/n = k-1 \). The integral becomes \( \int_0^{k-1} \frac{1}{1+x} dx \).

Answer 1

The Correct Option is B

The given limit is \( L = \lim_{n\to\infty} \sum_{r=1}^{(k-1)n} \frac{1}{n+r} \).
(Assuming this interpretation where the last denominator is \(kn\), implying \(n+r = kn \Rightarrow r=(k-1)n\)).
This can be written as a limit of a Riemann sum.
\[ L = \lim_{n\to\infty} \sum_{r=1}^{(k-1)n} \frac{1}{n(1+r/n)} = \lim_{n\to\infty} \frac{1}{n} \sum_{r=1}^{(k-1)n} \frac{1}{1+r/n} \] Let \( x = r/n \).
When \( r=1, x \approx 0 \).
When \( r=(k-1)n, x = (k-1)n/n = k-1 \).
And \( dx \approx 1/n \).
So the limit becomes an integral: \[ L = \int_{0}^{k-1} \frac{1}{1+x} dx \] \[ = [\log|1+x|]_{0}^{k-1} = \log|1+(k-1)| - \log|1+0| \] \[ = \log|k| - \log|1| = \log k - 0 = \log k \] (Since \(k \in N\), \(k \ge 1\), so \(|k|=k\)).
This matches option (2).
Alternative interpretation if the sum is \( \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{kn} \): This means the last term's denominator is literally \(kn\), not \(n+kn\).
If the terms are \( \frac{1}{n+1}, \frac{1}{n+2}, \dots, \frac{1}{n+(k-1)n} \).
Number of terms is \( (k-1)n \).
This is what was used above.
If the general term is \( \frac{1}{n+i} \) and the last term is \( \frac{1}{N_{final}} \) where \( N_{final}=kn \).
The sum is \( S_n = \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{kn} \).
The terms are of the form \( \frac{1}{n+r} \), where \( r \) goes from \(1\) to \( kn-n = (k-1)n \).
The calculation above is correct for this interpretation.