# You Can't Handle the Proof!

January 25, 2014 2:35 PM Subscribe

Working through Apostol's Calculus Volume I for personal development. Stuck on a proof. Can you help?

Trying to prove that 2(sqrt(n+1) - sqrt(n)) < 1/sqrt(n) < 2(sqrt(n) - sqrt(n-1)), n >= 1. I can make a damn fine geometric argument for this, but I can't seem to analytically prove it from the field, order, and least-upper bound axioms. Hope me!

Trying to prove that 2(sqrt(n+1) - sqrt(n)) < 1/sqrt(n) < 2(sqrt(n) - sqrt(n-1)), n >= 1. I can make a damn fine geometric argument for this, but I can't seem to analytically prove it from the field, order, and least-upper bound axioms. Hope me!

Also, good on you for doing calculus for personal development!

posted by number9dream at 3:04 PM on January 25, 2014

posted by number9dream at 3:04 PM on January 25, 2014

Ugh. I feel the dumb now. One of those times where you stare at something so long you can't see what's right in front of you. Anyway, thank you!

posted by bfranklin at 3:11 PM on January 25, 2014

posted by bfranklin at 3:11 PM on January 25, 2014

Are you using the 1st or 2nd edition?

In my opinion the 2nd has much harder problem sets and makes more of an effort to derive everything from first principles, sometimes to an excessive degree-- and I seem to remember that it begins with integration rather than the usual differentiation, though the 1st may, as well (my memory of that one is even dimmer).

posted by jamjam at 3:30 PM on January 25, 2014

In my opinion the 2nd has much harder problem sets and makes more of an effort to derive everything from first principles, sometimes to an excessive degree-- and I seem to remember that it begins with integration rather than the usual differentiation, though the 1st may, as well (my memory of that one is even dimmer).

posted by jamjam at 3:30 PM on January 25, 2014

I'm using the 2nd, and honestly, I'm happy to hear that it's harder and more from first principles. This isn't my first time through calc; I could probably do 90% of the problems in a more numeric calc book without too much difficulty. I'm really doing this to bring my proof skills back up to snuff (it's been 15 years) and make sure I understand things from first principles.

posted by bfranklin at 3:43 PM on January 25, 2014

posted by bfranklin at 3:43 PM on January 25, 2014

Here is another approach: Multiply both sides of the first inequality by

sqrt(n+1) + sqrt(n),

use the difference-of-squares identity (a-b)*(a+b) = a^2-b^2, and marvel at the results. This trick is often handy for clearing out radicals.

posted by aws17576 at 7:58 PM on January 25, 2014 [2 favorites]

sqrt(n+1) + sqrt(n),

use the difference-of-squares identity (a-b)*(a+b) = a^2-b^2, and marvel at the results. This trick is often handy for clearing out radicals.

posted by aws17576 at 7:58 PM on January 25, 2014 [2 favorites]

One way to come up with an inequality like this is to have the idea of writing n+1 as the square of something so you'll be able to take its square root; that might lead to the idea of completing the square:

n+1 = (sqrt n + 1/(2 sqrt n))^2 - 1/(4n)

Therefore

sqrt(n+1) < sqrt n +1/(2 sqrt n)

and rearranging gives the desired inequality.

posted by stebulus at 11:51 AM on January 26, 2014

n+1 = (sqrt n + 1/(2 sqrt n))^2 - 1/(4n)

Therefore

sqrt(n+1) < sqrt n +1/(2 sqrt n)

and rearranging gives the desired inequality.

posted by stebulus at 11:51 AM on January 26, 2014

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2(sqrt(n+1) - sqrt(n)) < 1/sqrt(n).

Divide both sides by 2 and multiply both sides by sqrt(n). Then we are trying to show that the (equivalent) inequality

sqrt(n(n+1)) - n < 1/2

holds. Add n to both sides, so the inequality becomes

sqrt(n(n+1)) < n + 1/2.

Then square both sides, so now what we are trying to show is that

n(n+1) < (n+1/2)^2.

Expanding both sides gives:

n^2 + n < n^2 + n + 1/4.

So you can work backwards from here to derive the original inequality that we wanted. The other inequality should work similarly.

posted by number9dream at 3:01 PM on January 25, 2014