Question:

<p><strong>Step 1:</strong> Rewrite the sum by factoring out 1/n explicitly:</p><p>$$\lim_{n \to \infty} \frac{1}{n} \sum_{r=n+1}^{2n} \log\left(1 + \frac{r}{n}\right) = \lim_{n \to \infty} \sum_{r=n+1}^{2n} \frac{1}{n}\log\left(1 + \frac{r}{n}\right)$$</p><p><strong>Step 2:</strong> Reindex by setting r = n+k where k goes from 1 to n:</p><p>$$= \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n}\log\left(1 + \frac{n+k}{n}\right) = \lim_{n \to \infty} \sum_{k=1}^{n} \frac{1}{n}\log\left(2 + \frac{k}{n}\right)$$</p><p><strong>Step 3:</strong> Alternatively, substitute x = r/n directly. As r goes from n+1 to 2n, x goes from 1 to 2, with Δx = 1/n:</p><p>$$= \int_{1}^{2} \log(1+x)\,dx$$</p><p><strong>Step 4:</strong> Integrate by parts with u = log(1+x), dv = dx:</p><p>$$\int \log(1+x)\,dx = x\log(1+x) - \int \frac{x}{1+x}\,dx = x\log(1+x) - x + \log(1+x) + C$$</p><p><strong>Step 5:</strong> Evaluate from 1 to 2:</p><p>$$\left[x\log(1+x) - x + \log(1+x)\right]_1^2 = \left[2\log 3 - 2 + \log 3\right] - \left[\log 2 - 1 + \log 2\right]$$</p><p>$$= 3\log 3 - 2 - 2\log 2 + 1 = 3\log 3 - 2\log 2 - 1$$</p><p>∴ Answer: <strong>3log 3 − 2log 2 − 1</strong> (or equivalently <strong>log 27 − log 4 − 1</strong>)</p>