(1) limn→∞(1n+1+1n+2+1n+3+⋯+1n+n)=limn→∞∑k=1n1n+k=limn→∞∑k=1n1n⋅11+kn\lim_{n\to\infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots+\frac{1}{n+n}\right)=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n+k}=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n}\cdot\frac{1}{1+\dfrac{k}{n}}limn→∞(n+11+n+21+n+31+⋯+n+n1)=limn→∞∑k=1nn+k1=limn→∞∑k=1nn1⋅1+nk1 これから =∫0111+x dx=[log∣1+x∣]01=log2=\int_0^1 \frac{1}{1+x}\,dx=\Big[\log|1+x|\Big]_0^1=\log 2=∫011+x1dx=[log∣1+x∣]01=log2 (2) limn→∞(nn2+1+nn2+4+nn2+9+⋯+nn2+n2)=limn→∞∑k=1nnn2+k2=limn→∞∑k=1n1n⋅11+(kn)2\lim_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{n}{n^2+k^2}=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{1}{n}\cdot\frac{1}{1+\left(\dfrac{k}{n}\right)^2}limn→∞(n2+1n+n2+4n+n2+9n+⋯+n2+n2n)=limn→∞∑k=1nn2+k2n=limn→∞∑k=1nn1⋅1+(nk)21 これから =∫0111+x2 dx=[tan−1x]01=π4=\int_0^1 \frac{1}{1+x^2}\,dx=\Big[\tan^{-1}x\Big]_0^1=\frac{\pi}{4}=∫011+x21dx=[tan−1x]01=4π