lim n → ∞ ( 1 + 1 2 + 1 3 + ⋯ + 1 n ) 1 n \lim\limits_{n\rightarrow\infty}(1 + \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n})^{\frac{1}{n}} n→∞lim(1+21+31+⋯+n1)n1
lim n → ∞ ( 1 n + 1 + 1 n + 2 + ⋯ + 1 n + n ) \lim\limits_{n\rightarrow\infty}(\frac{1}{n+\sqrt{1}}+\frac{1}{n+\sqrt{2}}+\cdots+\frac{1}{n+\sqrt{n}}) n→∞lim(n+11+n+21+⋯+n+n1)
lim n → ∞ ∑ k = n 2 ( n + 1 ) 2 1 k \lim\limits_{n\rightarrow\infty}\sum\limits_{k = n^2}^{(n + 1)^2}\frac{1}{\sqrt{k}} n→∞limk=n2∑(n+1)2k1
lim n → ∞ n lg n n \lim\limits_{n\rightarrow\infty}\sqrt[n]{n\lg n} n→∞limnnlgn
lim n → ∞ ( 1 2 + 3 2 2 + ⋯ + 2 n − 1 2 n ) \lim\limits_{n\rightarrow\infty}(\frac{1}{2}+\frac{3}{2^2}+\cdots+\frac{2n - 1}{2^n}) n→∞lim(21+223+⋯+2n2n−1)
已知 lim n → ∞ a n = a \lim\limits_{n\rightarrow\infty}a_{n}=a n→∞liman=a, lim n → ∞ b n = b \lim\limits_{n\rightarrow\infty}b_{n}=b n→∞limbn=b,证明: lim n → ∞ a 1 b n + a 2 b n − 1 + ⋯ + a n b 1 n = a b \lim\limits_{n\rightarrow\infty}\frac{a_{1}b_{n}+a_{2}b_{n - 1}+\cdots+a_{n}b_{1}}{n}=ab n→∞limna1bn+a2bn−1+⋯+anb1=ab