limit of a recursive sequence_1 
I have this sequence 
$$ \begin{cases}a_1= \alpha +2  & \alpha >0\\
a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1} & n \ge1\end{cases}$$
and I want to know 
  $\lim_\limits{n\rightarrow \infty} a_n$

I can prove by induction that the sequence is a monotone increasing.
$$
a_1= \alpha +2 
$$
$$
a_2= \alpha +\frac{3}{2}
$$
$$
a_3= \alpha^2+2\alpha +\frac{4}{3}
$$
$$
a_4= (\alpha^2+2\alpha +1)^3+\frac{1}{4}
$$
if we suppose that $$a_n>a_{n-1} \rightarrow a_n=\left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}+ \frac{1}{n}>a_{n-1} \rightarrow \left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}>a_{n-1}- \frac{1}{n}$$
Then, $$a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1}=a_{n+1}=\left(\left(a_{n-1}- \frac{1}{n-1}\right)^{n-1}+ \frac{1}{n}-\frac{1}{n}\right)^n+\frac{1}{n+1}=\left(a_{n-1}- \frac{1}{n-1}\right)^{n(n-1)}+\frac{1}{n+1}>\left(a_{n-1}- \frac{1}{n}\right)^{n}+\frac{1}{n+1}$$
$$a_{n+1}>\left(a_{n-1}- \frac{1}{n}\right)^{n}+\frac{1}{n+1}=a_n \rightarrow  a_{n+1}>a_n$$
We can say that $a_{n+1}>a_n>0$
$$\lim_{n\rightarrow \infty} a_n=\sigma$$ can be a $\alpha \in R$ or $\infty$.
In particular we can say that:
$$a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1} \rightarrow a_{n+1}- \frac{1}{n+1}=\left(a_n- \frac{1}{n}\right)^n  \rightarrow \sigma= \sigma^n \rightarrow \sigma=+ \infty$$
 A: You can try unfolding the recursion:
\begin{align}
a_{n+1} &= \left(a_n - \frac1n\right)^n + \frac{1}{n+1}\\
&= \left(a_{n-1} - \frac1{n-1}\right)^{(n-1)n} + \frac{1}{n+1}\\
&= \left(a_{n-2} - \frac1{n-2}\right)^{(n-2)(n-1)n} + \frac{1}{n+1}\\
&\cdots\\
&= \left(a_{2} - \frac1{2}\right)^{2\cdot3 \cdots(n-1)n} + \frac{1}{n+1}\\
&= \left(a_1 - 1\right)^{n!} + \frac{1}{n+1}\\
&= \left(\alpha +1\right)^{n!} + \frac{1}{n+1}\\
\end{align}
Letting $n \to\infty$ gives $\lim_{n\to\infty} a_{n+1} = +\infty$ because $\alpha + 1 > 1$.
A: This series diverges towards infinity as you noted. In this case, instead of a good old "$\varepsilon - \delta$" proof, we need a $K-\delta$ proof.
Note first that:
$$
a_n >\frac{n+1}{n} \Rightarrow a_{n+1}>2+\frac{1}{n+1}
$$
Because:
$$
a_{n+1} = \left(a_n-\frac{1}{n} \right)^n +\frac{n+2}{n+1}> \left(\frac{n+1}{n}-\frac{1}{n} \right)^n +\frac{n+2}{n+1}= (1)^n +\frac{n+2}{n+1}>2+\frac{1}{n+1}
$$
So $a_n>2$ for $\alpha>0$ and $n>2$.
Therefore:
$$
a_{n+1} >  \left(2+\frac{1}{n}-\frac{1}{n} \right)^n +\frac{n+2}{n+1}> 2^n>K
$$
So for all $n>\log_2(K)$, $a_n>K$, and hence $a_n$ diverges towards positive infinity.
