Prove series is divergent (sqrt(n+1) - sqrt(n))/sqrt(n)

In summary, the author is trying to find a solution to an equation involving limits, but is struggling because of RGVParentheses in his LaTeX.
  • #1
srl17
13
0

Homework Statement


Prove that [itex] \sum\limits_{n = 0}^\infty {\frac{{\left( { \sqrt{n+1} \right) - \sqrt{n} }}{{\left( {\sqrt{n}} \right)!}}}[/itex]
is divergent

Homework Equations


The Attempt at a Solution


This is an intro to analysis course. We haven't gone over the integral test which would be wonderful here. I have tried the limit comparison w/ 1/n^1/2, ratio and root test which were all inconclusive. I thought of using the comparison test but 1/n^(1/2) is bigger.
I am thinking of using Cauchy Criterion for Series and proving that the partial sums are monotone increasing and unbounded, but how would I prove it is unbounded?

Or If anyone sees a simpler way than Cauchy I am all eyes.
And this is my first attempt at using latex so I hope the equation turns out right, if not sorry and reference the subject title. Thank you!
 
Last edited:
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  • #2
Multiply numerator and denominator by sqrt(n+1)+sqrt(n) and simplify the numerator. Then think about a comparison again.
 
  • #3
Thanks! After using the conjugate the limit comparison with 1/sqrt(n) came out to 0 but when done with 1/n the limit came out to be 1/2 which is usable. Thanks!
 
  • #4
You can also do it by showing that sqrt(1+n)/sqrt(n) - 1 > 1/(2n) - 1/(8n^2) for n > 1. In turn, this can be done by looking at the function f(x) = sqrt(1+x) - 1 - x/2 + x^2/8: f is strictly convex (f''(x) > 0) on x > 0; since f(0) = 0 and f '(0) = 0, f is strictly increasing and positive on x > 0.

RGV
 
Last edited:
  • #5
Parentheses in your LaTeX are mismatched, on several levels.
 
  • #6
srl17 said:

Homework Statement


Prove that [itex]\sum_{n=0}^\infty \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}}[/itex] is divergent

A better and simpler way, which I have shown above, would be to use this LaTeX code. All those extra {}s and parentheses are useless, and the sum code was entered incorrectly.

\sum_{n=0}^\infty \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}}
 

What is a series?

A series is a mathematical expression that represents the sum of a sequence of numbers. It can be written in the form of ∑(n=1 to ∞) a(n), where a(n) is the nth term in the sequence.

What does it mean for a series to be divergent?

A divergent series is one that does not have a finite sum, meaning that the sum of the terms in the series grows infinitely large as the number of terms approaches infinity.

How can I tell if a series is divergent?

One way to determine if a series is divergent is to use the divergence test, which states that if the limit of the terms in the series does not approach zero, then the series is divergent.

How do I apply the divergence test to this series?

To apply the divergence test to the series (sqrt(n+1) - sqrt(n))/sqrt(n), we can rewrite the expression as (sqrt(n+1)/sqrt(n)) - 1. Then, we can take the limit as n approaches infinity, which would be equal to 1 - 1 = 0. Since the limit is equal to zero, the series is divergent.

Are there any other tests that can be used to determine if a series is divergent?

Yes, there are several other tests that can be used, such as the integral test, comparison test, and ratio test. However, for this specific series, the divergence test is the most efficient method to determine its divergence.

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