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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}$

  • 0
    @Chon I'm guessing by exact you mean closed form, right?2015-11-11

3 Answers 3

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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

So:

$ a<\sqrt{n}<\sqrt{n+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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    Your last sum should start at $2$, not $1$.2012-02-22
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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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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 $\frac{3}{4}$