Integer part of a sum (floor)
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!
Your sum is telescoping since
$$S={\displaystyle \sum_{k=1}^{2014}\frac{x_{k}}{x_{k+1}}=\sum_{k=1}^{2014}\frac{x_{k}^{2}}{x_{k}x_{k+1}}=\sum_{k=1}^{2014}\frac{x_{k+1}-x_{k}}{x_{k}x_{k+1}}=\sum_{k=1}^{2014}(\frac{1}{x_{k}}-\frac{1}{x_{k+1}})} $$ $$ {\displaystyle =\frac{1}{x_{1}}-\frac{1}{x_{2015}}=2014-\frac{1}{x_{2015}}.}$$
Hence it's a matter of evaluating $x_{2015}$. It seems that $.17<x_{2015}<.18$ and $S=2008.22$.
Some identities that might help $$ x_{2015}=x_1+\sum_{k=1}^{2014} x_k^2;\ x_{2015}=x_1(1+x_1)(1+x_2)\cdot\ldots\cdot(1+x_{2014})$$probably with some inequalities involving sums and products.
To get a bound on $\dfrac{1}{x_{2015}}$, let $y_n = \dfrac{1}{x_n}$. Then, $\dfrac{1}{y_{n+1}} = \dfrac{1}{y_n}+\dfrac{1}{y_n^2} = \dfrac{y_n+1}{y_n^2}$.
Hence, $y_{n+1} = \dfrac{y_n^2}{y_n+1} = y_n-\dfrac{y_n}{y_n+1}$. Rearrange to get $\left(1+\dfrac{1}{y_n}\right)(y_n-y_{n+1}) = 1$.
Since $y_n$ is a decreasing sequence and $1+\dfrac{1}{y}$ is a decreasing function, we have:
$2014 = \displaystyle\sum_{n = 1}^{2014}\left(1+\dfrac{1}{y_n}\right)(y_n-y_{n+1}) = \sum_{n = 1}^{2014}\int_{y_{n+1}}^{y_n}\left(1+\dfrac{1}{y_n}\right)\,dy \le \sum_{n = 1}^{2014}\int_{y_{n+1}}^{y_n}\left(1+\dfrac{1}{y}\right)\,dy = \int_{y_{2015}}^{y_1}\left(1+\dfrac{1}{y}\right)\,dy = \left[y + \ln y\right]_{y_{2015}}^{y_1} = (2014+\ln 2014)-(y_{2015}+\ln y_{2015})$
Therefore, $y_{2015}+\ln y_{2015} \le \ln 2014$. Exponentiation yields $y_{2015}e^{y_{2015}} \le 2014$.
If you can convince yourself that $6e^6 > 2014$, then we have $y_{2015} \le 6$ (because $ye^y$ is increasing).
Therefore, $S = 2014 - y_{2015} \ge 2008$. It remains to show that $S \le 2009$, i.e. $y_{2015} \ge 5$.
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.