# Convergence

• Oct 10th 2006, 08:17 AM
OntarioStud
Convergence
Let yn = the sqrt(n+1)-sqrt(n).
Prove that both yn and sqrt(n)yn converge.

Could someone help me?
• Oct 10th 2006, 08:56 AM
ThePerfectHacker
Quote:

Originally Posted by OntarioStud
Let yn = the sqrt(n+1)-sqrt(n).
Prove that both yn and sqrt(n)yn converge.

Could someone help me?

We note that,
{y_n} is a postive sequence because,
n+1>n thus, sqrt(n+1)>sqrt(n) thus, sqrt(n+1)-sqrt(n)>0
Thus it has a lower bound.

The sequence is also strictly decreasing.
Because,
y_{n+1}<y_n
If and only if,
y_{n+1}-y_n<0
If and only if,
sqrt(n+2)-sqrt(n+1)-sqrt(n+1)+sqrt(n)<0
If and only if,
sqrt(n+2)-sqrt(n)<2sqrt(n+1)
Which is true (square both sides).

Thus, by Weierstrauss-Bolzano Theorem this sequence has a limit.
• Oct 10th 2006, 09:02 AM
ThePerfectHacker
For the second problem:

We have,
y_n=sqrt(n+1)-sqrt(n)
Thus,
x_n=sqrt(n)y_n=sqrt(n^2+n)-n
We can show that,
x_{n+1}>x_n
Thus it is strictly increasing.

Then we show using induction that,
x_n<1

Again, it must converge.