# Why does the limit(n->∞) sqrt(n)/(sqrt(n)+sqrt(n+1)) equal 1/(1+sqrt(1-1/n))?

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1. Nov 5, 2014

### h0p3d545

1. The problem statement, all variables and given/known data
I am studying for a calculus test tomorrow on this website (http://archives.math.utk.edu/visual.calculus/6/index.html). I am working on the limit comparison test problems but I am unfamiliar with the form they use in their solutions. For example:
Limit comparison test (prove convergence / divergence)
Series (from n=1 to ∞) 1/(sqrt(n)+sqrt(n+1))

2. Relevant equations

3. The attempt at a solution
Series (from n=1 to ∞) 1/(sqrt(n)+sqrt(n+1)) compare to Series (from n=1 to ∞) 1/sqrt(n), which we know diverges by the p-series test (p=.5, p<1)
Let an = 1/(sqrt(n)+sqrt(n+1)) and bn = 1/sqrt(n)
lim(n->∞) an/bn
= lim(n->∞) 1/(sqrt(n)+sqrt(n+1)) / 1/sqrt(n)
= lim(n->∞) sqrt(n)/(sqrt(n)+sqrt(n+1))

I can see why this limit is 1/2 but I don't understand how the solution to the problem ended up in the form
lim(n->∞) 1/(1+sqrt(1-1/n)) = 1/2
I can tell by the other solutions that this form is a rule of sorts but I don't know what it is.
Thank you for the help.

2. Nov 5, 2014

### Staff: Mentor

Factor $\sqrt{n}$ from the two terms of $\sqrt{n}$ + $\sqrt{n + 1}$.

3. Nov 5, 2014

### h0p3d545

OH. so if you multiply
sqrt(n)/(sqrt(n)+sqrt(n+1)) by (1/sqrt(n))/(1/sqrt(n)) you get 1/(sqrt(n)+sqrt(n+1))/sqrt(n) which can simplify to 1/(sqrt(n/n)+sqrt(n/n+1/n)) or 1/(1+sqrt(1+1/n))

4. Nov 5, 2014

### Staff: Mentor

No, it's simpler than that. Just factor $\sqrt{n}$ out of the numerator and denominator. The $\sqrt{n}$ factors can be cancelled, and you're left with the form you see in the solution.

5. Nov 5, 2014

### h0p3d545

Oh, ok. I understand. Thank you!