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

  • Thread starter Thread starter srl17
  • Start date Start date
  • Tags Tags
    Divergent Series
srl17
Messages
13
Reaction score
0

Homework Statement


Prove that \sum\limits_{n = 0}^\infty {\frac{{\left( { \sqrt{n+1} \right) - \sqrt{n} }}{{\left( {\sqrt{n}} \right)!}}}
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:
Physics news on Phys.org
Multiply numerator and denominator by sqrt(n+1)+sqrt(n) and simplify the numerator. Then think about a comparison again.
 
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!
 
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:
Parentheses in your LaTeX are mismatched, on several levels.
 
srl17 said:

Homework Statement


Prove that \sum_{n=0}^\infty \frac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n}} 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}}
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top