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I would like to prove the existence and the exact value of the following series:

$$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}$$

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    $\begingroup$ Hint: When is $\lfloor\sqrt{n+1}\rfloor \neq \lfloor \sqrt{n}\rfloor$? $\endgroup$ – Thomas Andrews Feb 22 '12 at 14:58
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    $\begingroup$ Think about the first quite a few terms. It is easy to write them down explicitly. You should get something familiar, a famous series. $\endgroup$ – André Nicolas Feb 22 '12 at 15:06
  • $\begingroup$ @Chon I'm guessing by exact you mean closed form, right? $\endgroup$ – user285523 Nov 11 '15 at 4:35
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If $ n+1 $ is a square, $\lfloor \sqrt{n+1} \rfloor=\sqrt{n+1} $

$$ 0<\sqrt{n+1}-\sqrt{n}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<1$$

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

So: $$ \lfloor \sqrt{n} \rfloor= \sqrt{n+1}-1$$

$$ \lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor=1 $$

I have proved:

$n+1$ is a square $\Longrightarrow \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor$

Now I must show that

$ \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor \Longrightarrow $ $n+1$ is a square

If $ n+1$ is not a square:

If $ n $ is a square, $\lfloor \sqrt{n} \rfloor=\sqrt{n}$. As $$ \sqrt{n+1}-1<\sqrt{n}<\sqrt{n+1} $$

$ \lfloor \sqrt{n+1} \rfloor= \sqrt{n} $

So: $ \lfloor \sqrt{n+1} \rfloor= \lfloor \sqrt{n} \rfloor $

If $ n $ is not a square, there exists $a\in \mathbb{N} $ such that

$$ a^2<n<n+1<(a+1)^2$$

So:

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

So : $$ \lfloor \sqrt{n+1} \rfloor= \lfloor \sqrt{n} \rfloor $$

Finally:

$n+1$ is a square $\Longleftrightarrow \lfloor{\sqrt{n+1}}\rfloor \neq \lfloor{\sqrt{n}}\rfloor$

So: $$ \sum_{n=1}^{\infty} \frac{\lfloor{\sqrt{n+1}}\rfloor-\lfloor{\sqrt{n}}\rfloor}{n}=\sum_{n=2}^{\infty} \frac{1}{n^2-1}=\cdots=\frac{3}{4}$$

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  • $\begingroup$ Good. You could have saved some space. For example, if $n+1$ is not a square, then there exists an integer $a$ such that $a^2\le n<n+1<(a+1)^2$. So you don't need to consider $n$ a square, not a square separately. $\endgroup$ – André Nicolas Feb 22 '12 at 16:59
  • $\begingroup$ You are right, thank you! $\endgroup$ – Chon Feb 22 '12 at 17:14
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    $\begingroup$ Your last sum should start at $2$, not $1$. $\endgroup$ – David Mitra Feb 22 '12 at 17:26
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Work on Thomas Andrews's hint (That is a very good hint)

Let me give a following hint

$$ \left\lfloor \sqrt{(3+1)} \right\rfloor = 2, \hspace{3pt} \left\lfloor \sqrt{(2+1)} \right\rfloor = 1, \text{ why?} $$

$$ \left\lfloor \sqrt{(8+1)} \right\rfloor = 3, \hspace{3pt} \left\lfloor \sqrt{(7+1)} \right\rfloor = 2, \text{ why?} $$

Try working towards these observed values, and in general for what values are they not equal?

The answer is $$\frac{3}{4}$$

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As all terms of the given series $s=\sum_{n=1}^\infty a_n$ are $\geq0$ we may collect them in packets and write $$s=\sum_{r=1}^\infty\left(\sum\nolimits_{r^2\leq n<(r+1)^2} a_n\right)\ .$$ Note that in the inner sum only the last term, corresponding to $n=(r+1)^2-1$, is nonzero and has the value $${1\over(r+1)^2-1}={1\over (r+2) r}={1\over2}\Bigl({1\over r}-{1\over r+2}\Bigr)\ .$$ It follows that the outer sum is a telescoping series, and we obtain $$s={1\over 2}\sum_{r=1}^\infty \Bigl({1\over r}-{1\over r+2}\Bigr)={1\over2}\bigl(1+{1\over2}\Bigr)={3\over4}\ .$$

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