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Let $x_n = 1 + \frac{1}{2} + \dots + \frac{1}{n} - \left\lfloor 1 + \frac{1}{2} +\dots+\frac{1}{n}\right\rfloor \ \forall n \in \mathbb{N} $ be a sequence . Prove that it is not convergent?

$\lfloor x\rfloor$ means floor.

I have absolutely no ideea how to prove that is not convergent ??

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    Either there's some "obscure" trick or you miscopied the question: as far as I can see, you get a rather boring yet convergent sequence: $\;x_n=0\;,\;\;\forall\,n\in\Bbb N\;$ , and it's equally trivial to check the series whose sequnece of partial sums is the above one is *also* convergent, and it converges to zero... – 2017-02-03
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    @Ed Well, then it is boringly trivial. – 2017-02-03
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    This looks incomplete. What is the series? Is it $\sum_{n=1}^{\infty} x_n$? – 2017-02-03
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    @DonAntonio: if $[x]$ stands for the largest integer $\leq x$, or for the closest integer to $x$ (as I believe), the question is not entirely trivial. But this point should be clarified. – 2017-02-03
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    Does $[\,\cdot\,]$ denote the Gauß bracket, aka. greatest integer function, aka. floor? – 2017-02-03
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    @JackD'Aurizio If it is that, then yes. Yet after my first comment the OP said nothing about this...so I am not sure. – 2017-02-03
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    No no, i m sorry. It was a mistake after all. I was writing series instead of sequence. – 2017-02-03
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    But the big question is the definition of $\left[x\right]$. Do you mean "greatest integer" or not? – 2017-02-03
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    $[x]$ means floor. – 2017-02-03
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    The summation of 1/n is unbounded - can you show that? It might look bounded, but it isn't. By considering enough future terms, you can always add another 1/2 to the summation, if you add another n terms. After showing that, can you see what happens to x as n increases to infinity? It has to oscillate between 0 and 1, once the summation clocks past the next integer, the whole summation drops down to 0 or just above 0, then starts climbing back to wards 1 again - the whole question is non-trivial, it does require some work, yes! – 2017-02-03

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Using the standard notations $\{x\}=x-\lfloor x\rfloor$ for the fractional part and $H_n=\sum_{k=1}^{n}\frac{1}{n}$ for harmonic numbers, we are asked to prove that the sequence of fractional parts of harmonic numbers is not convergent. That can be proved, for instance, by showing that every element $\alpha\in(0,1)$ is an accumulation point for such a sequence. $\{H_n\}_{n\geq 1}$ is a slowly increasing sequence since $H_n=\log(n)+\gamma+O\left(\frac{1}{n}\right)$. If we choose $n$ as the closest integer to $e^{M-\gamma+\alpha}$, with $M$ being a huge natural number, then $\{H_n\}$ is arbitrarily close to $\alpha$ and we are done.

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Jack,s proof applies to any sequence defined by $x_n = s_n -\lfloor s_n \rfloor$ where $s_n =\sum_{k=1}^n t_k$ with $t_n >0, t_{n+1}

Then $x_n$ will be dense in $[0,1]$ as can be seen by noting that, for any integer $m>0$, once $t_n<1/m$, every value in $[0,1]$ will be within $1/m$ of a $t_{n+k}$.