12.4 Limit of Series 数列的极限 - 独中课程 | 线上补习

高二数学 | 高级数学

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12.4 Limit of Series 数列的极限

Evaluate the following limit.
计算下列各极限

[1]

$\lim\limits_{n\rightarrow\infty}\frac{1}{n^2}(1+2+3+...+n)$

Answer:

$\frac{1}{2}$

[2]

$\lim\limits_{n\rightarrow\infty}(\frac{1+2+3+...+n}{n+2}-\frac{n}{2})$

Answer:

$-\frac{1}{2}$

[3]

$\lim\limits_{n\rightarrow\infty}\frac{2+5+8+...+(3n-1)}{3+7+11+...+(4n+3)}$

Answer:

$\frac{3}{4}$

[4]

$\lim\limits_{n\rightarrow\infty}[\frac{1}{3n^2}\sum_{i=0}^{n}{(2i+1)}]$

Answer:

$\frac{1}{3}$

[5]

$\lim\limits_{n\rightarrow\infty}(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n(n+1)})$

Answer:

[6]

$\lim\limits_{n\rightarrow\infty}\frac{2^{n-1}-3^n}{2^n+3^{n-1}}$

Answer:

-3

[7]

$\lim\limits_{n\rightarrow\infty}(\frac{1}{n^2}+\frac{3}{n^2}+\frac{5}{n^2}+...+\frac{2(2n)-1}{n^2})$

Answer:

[8]

$\lim\limits_{n\rightarrow\infty}(\frac{1}{2\times4}+\frac{1}{3\times5}+\cdot\cdot\cdot+\frac{1}{(n+1)(n+3)})$

Answer:

$\frac{5}{12}$

[9]

$\lim\limits_{n\rightarrow\infty}(\frac{1}{1^2+3}+\frac{1}{2^2+6}+\cdot\cdot\cdot+\frac{1}{n^2+3n})$

Answer:

$\frac{11}{18}$

[10]

$\lim\limits_{n\rightarrow\infty}\frac{1+3+5+...+(2n-1)}{5+8+11+...+(3n-1)}$

Answer:

$\frac{2}{3}$

[11]

$\lim\limits_{n\rightarrow\infty}\frac{1^2+2^2+3^2+...+(n+1)^2}{2n^3-5n^2+1}$

Answer:

$\frac{1}{6}$

[12]

$\lim\limits_{n\rightarrow\infty}\frac{1^2+3^2+5^2+...+(2n-1)^2}{2^2+4^2+6^2+...+(2n)^2}$

Answer:

[13]

$\lim\limits_{n\rightarrow\infty}[(1-\frac{1}{2^2})(1-\frac{1}{3^2})\cdot\cdot\cdot(1-\frac{1}{n^2})(1-\frac{1}{(n+1)^2})]$

Answer:

$\frac{1}{2}$

[14]

If f (n) = 1 + 2 + 3 + … + n (n $\epsilon$ N), then $\lim\limits_{n\rightarrow\infty}\frac{[f(n)]^2}{f(n^2)}$ = ?

若f (n) = 1 + 2 + 3 + … + n (n $\epsilon$ N), 则 $\lim\limits_{n\rightarrow\infty}\frac{[f(n)]^2}{f(n^2)}$ = ?

Answer:

$\frac{1}{2}$