<s>[INST] Problem of finding the limit of reductive sequence Let $\lbrace u_n \rbrace$ be a sequence defined by : $u_1=1$, $u_{n+1}=u_{n}+\dfrac{u_{n}^{2}}{a}$ for a be a positive real constant.
Find : $\text{lim}_{n\rightarrow \infty}\left(\dfrac{u_1}{u_2}+...+\dfrac{u_n}{u_{n+1}}\right)$
If I denote $S_n$ by $\dfrac{u_1}{u_2}+...+\dfrac{u_n}{u_{n+1}}$ then $S_n$ is increasing sequence, but I could not prove that it is bounded above. May be  $\text{lim}_{n\rightarrow \infty} S_n =\infty$ ? 
Please help me to solve it. Thanks
 [/INST] Shortly after finishing the proof, I realized that it is sufficient to observe the following identity
$$ \sum_{k=1}^{n} \frac{u_k}{u_{k+1}} + \frac{a}{u_{n+1}} = a $$
and the fact that $u_{n} \to \infty$ as $n\to\infty$. Indeed,
$$ \frac{a}{u_n} = \frac{u_n + a}{u_n + \frac{u_n^2}{a}} = \frac{u_n + a}{u_{n+1}} = \frac{u_n}{u_{n+1}} + \frac{a}{u_{n+1}}$$
and since $u_1 = 1$, the identity follows. Then by noting that
$$ u_{n} \geq u_{n-1}^2 \geq \cdots \geq (u_2)^{2^{n-2}} \to \infty \quad \text{as} \quad n\to\infty, $$
the conclusion follows.
I leave my old and sophisticated proof, which served a motivation.

Finally I succeeded in proving my observation that
$$ \sum_{n=1}^{\infty} \frac{u_n}{u_{n+1}} = a.$$

Step 0. Notation

By the substitution $u_n = \dfrac{a}{x_n}$, we have
$$ x_1 = a, \qquad x_{n+1} = \frac{x_n^2}{1+x_n}, \qquad \frac{u_n}{u_{n+1}} = \frac{x_{n+1}}{x_n}.$$
We denote $x_n = x_n(a)$ whenever the dependence of the sequence $(x_n)$ on the variable $a$ is required. Then we denote the limit as
$$s(a) := \sum_{n=1}^{\infty} \frac{u_n}{u_{n+1}} = \sum_{n=1}^{\infty} \frac{x_{n+1}(a)}{x_{n}(a)} \tag{1}$$
Since each summand is non-negative, it either converges absolutely or diverges to $+\infty$.

Step 1. Convergence and simple estimation of $s(a)$

Let $g(x)$ and $h(x)$ be functions defined on $x > 0$ by
$$ g(x) = \frac{x^2}{1+x} \quad \text{and} \quad h(x) = \frac{g(x)}{x} = \frac{x}{1+x}.$$
One the one hand, since $0 < h(x) < 1$,
$$ x_{n+1} = h(x_n) x_n < x_n$$
and $(x_n)$ decreases and converges to $0$ as $n \to \infty$. Then from the observation that $h(x)$ is an increasing function, we have
$$ x_{n+1} = h(x_n) x_n \leq h(x_1) x_n.$$
Inductively applying this inequality we obtain
$$ x_n \leq h(x_1)^{n-1} x_1 = \left(\frac{a}{1+a}\right)^{n-1} a. $$
On the other hand, 
$$ x_{n+1} \leq x_{n}^2 \leq \left(\frac{a}{1+a}\right)^{n-1} a x_n.$$
Therefore we have
$$ s(a) \leq \sum_{n=1}^{\infty} \left(\frac{a}{1+a}\right)^{n-1} a = a + a^2.$$
In particular, the defining summation $(1)$ converges absolutely. Since
$$ s(a) \geq \frac{x_2(a)}{x_1(a)} = \frac{a}{1+a} \geq a - a^2, $$
we obtain the following estimate:
$$ \left| s(a) - a \right| \leq a^2. \tag{2}$$

Step 2. Further estimation

Since $x_{n+1} = g(x_n)$, we have $x_{n+1}(a) = x_{n}(g(a))$. This gives the following identity:
$$ s(a) = \frac{x_{2}(a)}{x_{1}(a)} + \sum_{n=1}^{\infty} \frac{x_{n+1}(g(a))}{x_{n}(g(a))} = \frac{a}{1+a} + s\left(\frac{a^2}{1+a}\right). \tag{3}$$
Now we claim the following estimation:
$$ \left| s(a) - a \right| \leq a^{2^n} \quad \text{for all} \quad a > 0 \text{ and } n \geq 1. \tag{4} $$
The case $n = 1$ reduces to $(2)$, hence is true for the initial case $n = 1$. Thus assume that $(4)$ holds for $n$. By $(3)$,
$$\left| s(a) - a \right|
= \left| \frac{a}{1+a} + s\left(\frac{a^2}{1+a}\right) - a \right|
= \left| s\left(\frac{a^2}{1+a}\right) - \frac{a^2}{1+a} \right|
\leq \left(\frac{a^2}{1+a}\right)^{2^n}
\leq a^{2^{n+1}}.$$
Therefore $(4)$ holds for $n+1$, and consequently for all $n$ by mathematical induction.

Step 3. Proof of the main claim

Now we claim that $s(a) = a$ for all $a > 0$. Let us first consider the case $a \in (0, 1)$. Then taking $n \to \infty$ to the estimation $(4)$, we have $s(a) = a$.
Now we consider the case $a \geq 1$. Then note that $(3)$ can be written as
$$ s(a) = \frac{x_2(a)}{x_1(a)} + s(x_2(a)). $$
Thus inductively applying this identity, we obtain
$$ s(a) = \sum_{k=1}^{m-1} \frac{x_{k+1}(a)}{x_k(a)} + s(x_m(a)).$$
But we know that $x_m(a) < 1$ for large $m$. Thus with such $m$, the previous result implies
$$ s(a) = \sum_{k=1}^{m-1} \frac{x_{k+1}(a)}{x_k(a)} + x_m(a).$$
But since
$$ \frac{x_{k+1}}{x_k} + x_{k+1} = h(x_k) + g(x_k) = x_k,$$
the above sum reduces to $s(a) = a$ as desired, completing the proof. ////
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