Analysis Limit Theorems

• Mar 5th 2009, 10:48 AM
noles2188
Analysis Limit Theorems
Prove that lim[(1/n)-(1/(n-1))] = 0.

Prove that lim[(n+2)/((n^2)-3)] = 0.--for this one I know I need to use the definition that says a sequence (sn) is said to converge to the real number s provided that for each epsilon>0 there exists a real number M such that for all n elements of N, n>M implies that |(sn)-s|<epsilon
• Mar 12th 2009, 08:01 AM
GaloisTheory1
Quote:

Originally Posted by noles2188
Prove that lim[(1/n)-(1/(n-1))] = 0.

Prove that lim[(n+2)/((n^2)-3)] = 0.--for this one I know I need to use the definition that says a sequence (sn) is said to converge to the real number s provided that for each epsilon>0 there exists a real number M such that for all n elements of N, n>M implies that |(sn)-s|<epsilon

I am assuming this is:

$\displaystyle \lim_{n \rightarrow \infty}\frac{1}{n}-\frac{1}{n-1} = 0.$

$\displaystyle \lim_{n \rightarrow \infty}\frac{n-1 - n}{n(n-1)}$

$\displaystyle \lim_{n \rightarrow \infty}\frac{-1}{n(n-1)}$

$\displaystyle \lim_{n \rightarrow \infty}\frac{-1}{n^2-n}$

$\displaystyle \lim_{n \rightarrow \infty}\frac{\frac{-1}{n^2}}{1-\frac{1}{n}}$

do you see what happens now?
• Mar 12th 2009, 09:38 AM
ThePerfectHacker
Quote:

Originally Posted by noles2188
Prove that lim[(1/n)-(1/(n-1))] = 0.

What we basically want to show is that $\displaystyle \left| \tfrac{1}{n} - \tfrac{1}{n-1} \right|$ can be made arbitrary small for $\displaystyle n$ sufficiently large. Notice that $\displaystyle \left|\tfrac{1}{n} - \tfrac{1}{n-1}\right| = \tfrac{1}{n-1} - \tfrac{1}{n} = \tfrac{1}{n(n-1)} \leq \tfrac{1}{n}$. To complete the proof for $\displaystyle \epsilon > 0$ pick $\displaystyle N > \tfrac{1}{\epsilon}$ and so if $\displaystyle n > N$ then the definition of convergence is satisfied.