TITLE: Integer part of a sum (floor)
QUESTION [16 upvotes]: Let $\left(\, x_{n}\,\right)_{\,n\ \geq\ 1}$ be a sequence defined as follows:
$$
x_{1}={1 \over 2014}\quad\mbox{and}\quad x_{n + 1}=x_{n} + x_{n}^{2}\,,
\qquad\forall\ n\ \geq\ 1
$$
Compute the integer part of the sum:
$$
S=\frac{x_1}{x_2} + \frac{x_2}{x_3} + \cdots +\frac{x_{2014}}{x_{2015}}\,,\qquad
\left(\,\mbox{i. e.}\  \left\lfloor\, S\,\right\rfloor\,\right)$$
Any nice idea to approach this? How can one find a formula for $x_{n}$?
Thank you!

REPLY [5 votes]: JimmyK has ignored the request (in now deleted comments) to add the lower bound proof to his answer, so I will add an answer with the lower bound proof which will resolve the question completely.
Just like Jimmy's excellent upper bound proof we use
$$ 2014 = \sum_{n=1}^{2014}\left(1 + \frac{1}{y_n}\right)(y_n - y_{n+1}) = \sum_{n=1}^{2014} \int_{y_{n+1}}^{y_n} \left(1 + \frac{1}{y_n}\right) dy$$
Now we can easily show by induction that $y_k \gt -1$ and thus
$$y_{n+1} + 1 \gt y_{n+1} + \frac{y_n}{1+y_n} = y_n$$
Thus we have that
$$\left(1 + \frac{1}{y_n}\right) \ge \left(1 + \frac{1}{y_{n+1} +1}\right)$$
Now for $-1 \lt y_{n+1} \le y$ we have that
$$\left(1 + \frac{1}{y_{n+1} +1}\right) \ge \left(1 + \frac{1}{y +1}\right)$$
Thus
$$\left(1 + \frac{1}{y_n}\right) \ge \left(1 + \frac{1}{y +1}\right)$$
Let $a = y_{2015}$
We thus have
$$ 2014 = \sum_{n=1}^{2014} \int_{y_{n+1}}^{y_n} \left(1 + \frac{1}{y_n}\right)dy \ge \sum_{n=1}^{2014} \int_{y_{n+1}}^{y_n} \left(1 + \frac{1}{y +1}\right) dy = \int_{a}^{2014} \left(1 + \frac{1}{y +1}\right) dy =$$
$$ 2014-a + \log(2015) - \log(a+1)$$
Thus
$$ 2014 \ge 2014-a + \log(2015) - \log(a+1)$$
which means
$$ a + \log(a+1) \ge \log (2015)$$
giving us
$$ (a+1)e^{a} \ge 2015$$
Now $6e^5 \lt 6 \times3^5 = 1458$ and since $(x+1)e^x$ is increasing, we have $$a \gt 5$$
This coupled with the other two answers gives the answer of $$2008$$

An observation:
$$y_{n+1} = y_n - \frac{f(y_n)}{f'(y_n)}$$
where  $f(y) = ye^y$.
This is a Newton Raphson recurrence and techniques used there should be applicable here.