Let yn = the sqrt(n+1)-sqrt(n).

Prove that both yn and sqrt(n)yn converge.

Could someone help me?

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- Oct 10th 2006, 09:17 AMOntarioStudConvergence
Let yn = the sqrt(n+1)-sqrt(n).

Prove that both yn and sqrt(n)yn converge.

Could someone help me? - Oct 10th 2006, 09:56 AMThePerfectHacker
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, 10:02 AMThePerfectHacker
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.