Prove Squeezing Theorem for n Natural Numbers

[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.

 

[hypermonkey2]

ill post solution in the morning.

 

[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.

 

[devious_]

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


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

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

Second one should be similar.

 

[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. here's 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.

 

[hypermonkey2]

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

 

[hypermonkey2]

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

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

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

 

[hypermonkey2]

no takers? i post solutions tomorrow.

 

[redkimchi]

Yes, my solution ressembled the first one. Although, i was very amused at seeing the MVT making such short work of it. nice job. here's 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.

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

 

[redkimchi]

Finally, here's the cherry on top. Prove that the integer part of the following expression is odd.
$$(\sqrt{n}+\sqrt{n+1})^2$$
have fun! I am interested in seeing some original solutions as i think there are better ones than the ones i have seen.

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.

 

[CRGreathouse]

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.

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

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

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

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

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

 

[hypermonkey2]

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

 

[hypermonkey2]

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.

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?

 

[CRGreathouse]

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?

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

 

[hypermonkey2]

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

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

 

[Hurkyl]

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

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

 

[devious_]

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

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 $n < n+1$.

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

So we have:

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

 

[hypermonkey2]

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

If you say so, i haven't finished my cal 2 yet unfortunately, hehe. but ill keep it in mind!

 

[hypermonkey2]

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 $n < n+1$.
$$n < n+1<br /> \Rightarrow n^2 < n(n+1)<br /> \Rightarrow n < \sqrt{n(n+1)}<br /> \Rightarrow 2n - 2\sqrt{n(n+1)} < 0<br /> \Rightarrow n + 1 - 2\sqrt{n(n+1)} + n < 1<br /> \Rightarrow (\sqrt{n+1} + \sqrt{n})^2 < 1$$
So we have:
$$0 < \sqrt{n+1} - \sqrt{n} < 1<br /> \Rightarrow -1 < \sqrt{n} - \sqrt{n+1} < 0$$

oustanding, thanks for that.