# Fun math proof

1. Nov 12, 2005

### hypermonkey2

Any nice proofs for this?

$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$$

I hope the tex came out alright. have fun!

ps. n is any natural number.

Last edited: Nov 12, 2005
2. Nov 12, 2005

### hypermonkey2

ill post solution in the morning.

3. Nov 12, 2005

### daniel_i_l

I doesn't look that hard if you prove each half by induction, and if first you multiply everything by sqrt(n)/sqrt(n). But then again there could be a catch somewere. If I have time I'll try it.

4. Nov 12, 2005

### devious_

\begin{align*} 2\sqrt{n+1} - 2\sqrt{n} &= 2(\sqrt{n+1} - \sqrt{n}) \cdot \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}} \\ &= \frac{2}{\sqrt{n+1} + \sqrt{n}} \\ &< \frac{2}{\sqrt{n} + \sqrt{n}} \\ &= \frac{1}{\sqrt{n}} \end{align*}

Alternatively, let $f(x) = \sqrt{x}$, then by the MVT on [n, n+1]:

\begin{align*} \frac{\sqrt{n+1} - \sqrt{n}}{1} &= f'(x) \\ &= \frac{1}{2\sqrt{x}}, \text{ with x} \in (n, n+1) \\ &< \frac{1}{2\sqrt{n}} \end{align*}

Second one should be similar.

Last edited: Nov 12, 2005
5. Nov 13, 2005

### hypermonkey2

Yes, my solution ressembled the first one. Although, i was very amused at seeing the MVT making such short work of it. nice job. heres the pickle. i forgot the solution to this, ill think about it over the next few days, but have a try:
Using this inequality, evaluate to the $$\pm 1$$ the integer part of the sum
$$1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{1000000}}$$
any progress, you let me know.

6. Nov 13, 2005

### hypermonkey2

hehe, alright, it came back to me, but take a try at it anyways.

7. Nov 15, 2005

### hypermonkey2

Finally, heres the cherry on top. Prove that the integer part of the following expression is odd.

$$(\sqrt{n}+\sqrt{n+1})^2$$

have fun! im interested in seeing some original solutions as i think there are better ones than the ones i have seen.

Last edited: Nov 15, 2005
8. Nov 18, 2005

### hypermonkey2

no takers? i post solutions tomorrow.

9. Nov 18, 2005

### redkimchi

I think some appropriate definite integration of 1/sqrt(x) will give the desired approximation.

10. Nov 18, 2005

### redkimchi

This follows from the following 2 observations:
1.
$$(\sqrt{n}+\sqrt{n+1})^2 + (\sqrt{n}-\sqrt{n+1})^2 = 4n+2$$
2.
$$(\sqrt{n}-\sqrt{n+1})^2$$ is really small.

11. Nov 19, 2005

### CRGreathouse

$$2\sum_{k=a}^b\left(\sqrt{k+1}-\sqrt{k}\right)<\sum_{k=a}^b\frac{1}{\sqrt k}<2\sum_{k=a}^b\left(\sqrt{k}-\sqrt{k-1}\right)$$

$$2\left(\sqrt{b+1}-\sqrt{a}\right)<\sum_{k=a}^b\frac{1}{\sqrt k}<2\left(\sqrt{b}-\sqrt{a-1}\right)$$

$$2\left(\sqrt{1000001}-\sqrt{1}\right)<\sum_{k=1}^{1000000}\frac{1}{\sqrt k}<2\left(\sqrt{1000000}-\sqrt{0}\right)$$

$$1998<\sum_{k=1}^{1000000}\frac{1}{\sqrt k}<2000$$

By direct calculation
$$1998.5401<\sum_{k=1}^{1000000}\frac{1}{\sqrt k}<1998.5402$$

Last edited: Nov 19, 2005
12. Nov 19, 2005

### hypermonkey2

well done, youve saved me a bit of typing! yes thats the solution i have as well.

13. Nov 19, 2005

### hypermonkey2

interesting, but i find your second premise a little weak, as in... not justified. it definitely is true though. i could be wrong though, its just what if n is very large?

14. Nov 19, 2005

### CRGreathouse

It decreases strictly as n increases. In particular, if n>0, $$-1<\sqrt n-\sqrt{n+1}<0$$. This gives the desired result.

15. Nov 19, 2005

### hypermonkey2

interesting! im curious, is there a nice proof of this inequality?

16. Nov 19, 2005

### Hurkyl

Staff Emeritus
The original problem looks like an easy application of Taylor series! The 0-th order approximation is:

$$f(x + e) = f(x) + f'(x^*) e$$

for some $x^*$ between x and x+e inclusive.

Letting f be the square root function and e be 1, we have:

$$\sqrt{x + 1} - \sqrt{x} = \frac{1}{2 \sqrt{x^*}} \leq \frac{1}{2 \sqrt{x}}$$

because the rightmost term is an upper bound for the middle term. Letting e = -1, we have:

$$\sqrt{x - 1} - \sqrt{x} = - \frac{1}{2 \sqrt{x^*}} \geq -\frac{1}{2 \sqrt{x}}$$

because the rightmost term is a lower bound for the middle term. (Note the signs!)

Therefore,

$$\sqrt{x+1} - \sqrt{x} \leq \frac{1}{2 \sqrt{x}} \leq \sqrt{x} - \sqrt{x - 1}$$

or

$$2\sqrt{x+1} - 2\sqrt{x} \leq \frac{1}{\sqrt{x}} \leq 2\sqrt{x} - 2\sqrt{x - 1}$$

Last edited: Nov 19, 2005
17. Nov 19, 2005

### devious_

Yes. It's actually quite easy to see why this is true, since the terms in the middle are consecutive square roots. A formal proof goes something like this:

$$\sqrt{n+1} - \sqrt{n} > 0 \text{ is obviously true for } n \in \mathbb{N}$$

It remains to show that:

$$\sqrt{n+1} - \sqrt{n} < 1$$

Squaring both sides and playing around, we see that this follows from $n < n+1$.

$$n < n+1 \Rightarrow n^2 < n(n+1) \Rightarrow n < \sqrt{n(n+1)} \Rightarrow 2n - 2\sqrt{n(n+1)} < 0 \Rightarrow n + 1 - 2\sqrt{n(n+1)} + n < 1 \Rightarrow (\sqrt{n+1} + \sqrt{n})^2 < 1$$

So we have:

$$0 < \sqrt{n+1} - \sqrt{n} < 1 \Rightarrow -1 < \sqrt{n} - \sqrt{n+1} < 0$$

18. Nov 19, 2005

### hypermonkey2

If you say so, i havent finished my cal 2 yet unfortunately, hehe. but ill keep it in mind!

19. Nov 19, 2005

### hypermonkey2

oustanding, thanks for that.