[limite] Sa se arate ca:

Lauryca Lun Noi 19, 2012 10:12 pm

Sa se arate ca:
a)lim n^2/n+2=+infinit
b)lim [1*2+2*3+...+n(n+1)]/[n(1+3+5+...+(2n-1)]=1/3

**Admin** Lun Noi 19, 2012 10:21 pm

a)lim n^2/n+2=+infinit

Factor comun n ....

b)lim [1*2+2*3+...+n(n+1)]/[n(1+3+5+...+(2n-1)]=1/3

$\begin{array}{l} \mathop {\lim }\limits_{n \to \infty } \frac{{1 \times 2 + 2 \times 3 + \cdots + n\left( {n + 1} \right)}}{{n\left( {1 + 3 + 5 + \cdots + 2n - 1} \right)}}\\ \bullet 1 \times 2 + 2 \times 3 + \cdots + n\left( {n + 1} \right) = \sum\limits_{k = 1}^n {k\left( {k + 1} \right)} = \sum\limits_{k = 1}^n {{k^2}} + \sum\limits_{k = 1}^n k \\ = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} + \frac{{n\left( {n + 1} \right)}}{2} = \frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}.\\ \bullet 1 + 3 + 5 + \cdots + 2n - 1 = \sum\limits_{k = 1}^n {2k - 1} = 2\sum\limits_{k = 1}^n k - \sum\limits_{k = 1}^n 1 = 2 \times \frac{{n\left( {n + 1} \right)}}{2} - n = {n^2} \end{array}$
$\mathop {\lim }\limits_{n \to \infty } \frac{{1 \times 2 + 2 \times 3 + \cdots + n\left( {n + 1} \right)}}{{n\left( {1 + 3 + 5 + \cdots + 2n - 1} \right)}} = \mathop {\lim }\limits_{n \to \infty } \frac{{\frac{{n\left( {n + 1} \right)\left( {n + 2} \right)}}{3}}}{{n \times {n^2}}} = \frac{1}{3}$

[limite] Sa se arate ca:

[limite] Sa se arate ca:

Re: [limite] Sa se arate ca: