Technical question regarding showing sqrt(n+1) - sqrt(n) converges to 0

  • Thread starter gmn
  • Start date
In summary, the sequence sn= (n+1)1/2 - n1/2 converges to zero, with a tighter upper bound of 1/2(n1/2) compared to the initial upper bound of 1/n(1/2). Both bounds are true and sufficient to show the convergence to zero, but the second bound is preferred by the professor for being smaller. This can be seen through the rationalization of the numerator and using the fact that (n+1)^{1/2} > n^{1/2} in the attempt at a solution.
  • #1
gmn
4
0

Homework Statement



show the sequence sn= (n+1)1/2 - n1/2 converges to zero

Homework Equations





The Attempt at a Solution



I don't have that much of a problem showing the limit goes to zero, rationalize the numerator (or whatever it's called) to get (n+1)1/2 - n1/2 = 1/((n+1)1/2 + n1/2). My question is that I show this goes to zero because sn<1/n(1/2) which goes to zero, but my professor provides a solution where he writes sn<1/2(n1/2). I don't understand why the 2 is there. Is saying that sn<1/(n1/2) insufficient or not true?

Thanks
 
Physics news on Phys.org
  • #2
both are true, and sufficient to show it converges to zero, the 2nd is just a little tighter

how about this, as
[tex] (n+1)^{1/2} > n^{1/2} [/tex]
then
[tex] \frac{1}{s_n} = n^{1/2} + (n+1)^{1/2} > n^{1/2} + n^{1/2} = 2n^{1/2} [/tex]
then inverting
[tex] s_n = \frac{1}{n^{1/2} + (n+1)^{1/2}} < \frac{1}{n^{1/2} + n^{1/2}} = \frac{1}{2n^{1/2}} [/tex]
 
  • #3
[tex] \sqrt{n+1}+\sqrt{n}> 2 \sqrt{n}[/tex].
taking the inverse on both sides yields
[tex]\frac{1}{\sqrt{n+1}+\sqrt{n}} < \frac{1}{2 \sqrt{n}}[/tex]
your professor is just using a smaller upper bound for [tex]s_n[/tex]. professors like to use bounds that are as small as possible:smile:
 

1. What is the mathematical expression for showing sqrt(n+1) - sqrt(n) converges to 0?

The mathematical expression for this statement is lim(n->infinity) sqrt(n+1) - sqrt(n) = 0. This means that as n approaches infinity, the difference between the square roots of n+1 and n will approach 0.

2. How do you prove that sqrt(n+1) - sqrt(n) converges to 0?

To prove this statement, we can use the limit definition of convergence. This involves showing that for any given value of epsilon (ε > 0), there exists a value of n such that for all n > N, the expression sqrt(n+1) - sqrt(n) is within ε of 0. This can be demonstrated using algebraic manipulations and the concept of the squeeze theorem.

3. Can you provide an example of when sqrt(n+1) - sqrt(n) converges to 0?

One example of when this statement holds true is when n = 100. In this case, sqrt(100+1) - sqrt(100) = 10.05 - 10 = 0.05, which is very close to 0. As n increases, the difference between the square roots will continue to approach 0.

4. Are there any special cases where sqrt(n+1) - sqrt(n) does not converge to 0?

Yes, there are some special cases where this statement may not hold true. For example, if n is a negative number, the expression sqrt(n+1) - sqrt(n) would not be defined. Additionally, if n is a non-real complex number, the concept of convergence to 0 may not apply. It is important to consider the domain and limitations of the expression when evaluating its convergence.

5. How does this statement relate to other mathematical concepts?

This statement is related to the concept of sequences and their convergence. It also involves the use of algebraic manipulation and the squeeze theorem, which are important tools in mathematical analysis. Additionally, this statement can be seen as an application of the limit definition of convergence, which is a fundamental principle in calculus.

Similar threads

  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
301
  • Calculus and Beyond Homework Help
Replies
1
Views
251
  • Calculus and Beyond Homework Help
Replies
8
Views
659
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
986
  • Calculus and Beyond Homework Help
Replies
5
Views
481
  • Calculus and Beyond Homework Help
Replies
3
Views
544
  • Calculus and Beyond Homework Help
Replies
4
Views
811
Back
Top