(1) limn→∞∑k=1nkn(n+k)=limn→∞1n∑k=1nkn1+kn=∫01x1+x dx=∫01(1−11+x)dx=[x−log∣x+1∣]01=1−log2\lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{n(n+k)} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \frac{\dfrac{k}{n}}{1+\dfrac{k}{n}} = \int_0^1 \frac{x}{1+x}\,dx = \int_0^1 \left(1 - \frac{1}{1+x}\right)dx = \left[x - \log|x+1|\right]_0^1 = 1 - \log 2limn→∞∑k=1nn(n+k)k=limn→∞n1∑k=1n1+nknk=∫011+xxdx=∫01(1−1+x1)dx=[x−log∣x+1∣]01=1−log2 (2) limn→∞1n2∑k=1nn2−k2=limn→∞1n∑k=1n1−k2n2=∫011−x2 dx=[12(x1−x2+sin−1x)]01=π4\lim_{n \to \infty} \frac{1}{n^2} \sum_{k=1}^{n} \sqrt{n^2 - k^2} = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \sqrt{1 - \frac{k^2}{n^2}} = \int_0^1 \sqrt{1-x^2}\,dx = \left[\frac{1}{2}\left(x\sqrt{1-x^2} + \sin^{-1} x\right)\right]_0^1 = \frac{\pi}{4}limn→∞n21∑k=1nn2−k2=limn→∞n1∑k=1n1−n2k2=∫011−x2dx=[21(x1−x2+sin−1x)]01=4π