Does $\sqrt{n!+n}-\sqrt{n!} > 1$ for Some Integer n?

[Treadstone 71]

Is there an integer n such that

\sqrt{n!+n}-\sqrt{n!} > 1

?

 

[Hurkyl]

What have you tried?

*Sigh* If you really have no clue at all where to begin, try approximating things.

 

[arildno]

Congregrate, consummate, conjugate, conflagrate and contemplate.

 

[Cexy]

You may find that the approximation

\sqrt{n!+n} = \sqrt{n!}\sqrt{1+n/n!} = \sqrt{n!}\,\Big(1+\frac{1}{2(n-1)!}+\cdots\Big)

is useful.

 

[topsquark]

Is there an integer n such that

\sqrt{n!+n}-\sqrt{n!} > 1

?

Your function appears to have an absolute maximum (for positive integers, anyway) somewhere between n=1 and n=3 (near n=2 perhaps). The value of the function at n=2 is 2- \sqrt2 which is about 0.585786 <1. So I would say no. Unless n is negative. I haven't a clue how to calculate it in that case.

-Dan

 

[Treadstone 71]

Yes there is indeed a max < 1, after which it tends to 0, very, very rapidly.

Too bad, I was hoping to disprove Andrica's conjecture.

 

[arildno]

The simplest way to see this, is to first multiply with the conjugate expression:
\sqrt{n!+n}-\sqrt{n!}=\frac{(\sqrt{n!+n}+\sqrt{n!})(\sqrt{n!+n}-\sqrt{n!})}{\sqrt{n!+n}+\sqrt{n!}}\leq\frac{n}{2\sqrt{n!}}
One can easily prove the proposition from this bound:
For any n>3, we have
\frac{n+1}{2\sqrt{(n+1)!}}=\frac{n}{2\sqrt{n!}}\frac{1+\frac{1}{n}}{\sqrt{n+1}}&lt;\frac{n}{2\sqrt{n!}}\frac{2}{2}=\frac{n}{2\sqrt{n!}}
I.e, the sequence is decreasing.

 

[VietDao29]

The simplest way to see this, is to first multiply with the conjugate expression:
\sqrt{n!+n}-\sqrt{n!}=\frac{(\sqrt{n!+n}+\sqrt{n!})(\sqrt{n!+n}-\sqrt{n!})}{\sqrt{n!+n}+\sqrt{n!}}\leq\frac{n}{2\sqrt{n!}}
One can easily prove the proposition from this bound:
For any n>3, we have
\frac{n+1}{2\sqrt{(n+1)!}}=\frac{n}{2\sqrt{n!}}\frac{1+\frac{1}{n}}{\sqrt{n+1}}&lt;\frac{n}{2\sqrt{n!}}\frac{2}{2}=\frac{n}{2\sqrt{n!}}
I.e, the sequence is decreasing.

I am not very sure that I understand your post correctly, but do you mean:
\sqrt{(n + 1)! + (n + 1)} - \sqrt{(n + 1)!} \leq \frac{n + 1}{2 \sqrt{(n + 1)!}} &lt; \frac{n}{2 \sqrt{n!}} \geq \sqrt{n!+n}-\sqrt{n!}. And you conclude that:
\sqrt{(n + 1)! + (n + 1)} - \sqrt{(n + 1)!} \leq \sqrt{n! + n} - \sqrt{n!}.
Is that reasonable?
Am I missing something?

 

[arildno]

Hmm, no.
Multiplying out the numerator, we have the identity:
\sqrt{n!+n}-\sqrt{n!}=\frac{n}{\sqrt{n!+n}+\sqrt{n!}}\leq\frac{n}{2\sqrt{n!}}
This holds for ANY n.
Thus, if you can show that this bound is always less than 1 for any n, then it follows that your original difference expression must also be less than 1 for any n.

 

[VietDao29]

Hmm, no.
Multiplying out the numerator, we have the identity:
\sqrt{n!+n}-\sqrt{n!}=\frac{n}{\sqrt{n!+n}+\sqrt{n!}}\leq\frac{n}{2\sqrt{n!}}
This holds for ANY n.
Thus, if you can show that this bound is always less than 1 for any n, then it follows that your original difference expression must also be less than 1 for any n.

Whoops, srry for misinterpreting your post.
Should've read it more carefully.