Limit Question With Super Exponential Functions I'm given with the following sequence:
$ a_1 = \dfrac{\mathrm{e}^2+1}{\mathrm{e}^{\mathrm{e}^2}}, a_2 = \dfrac{\mathrm{e}^{\mathrm{e}^2+1}}{\mathrm{e}^{\mathrm{e}^{\mathrm{e}^2}}},\ldots$ 
It continues the same way (the numerator has $n$ powers of $\mathrm{e}$ , and the denominator $n+1 $ powers of $e$ ) . 
Does this sequence converges to zero? How can I prove this? 
If I want to upgrade the question:
What if I add a another term to the numerator, such as $\mathrm{e}^n $ ? 
The sequence will then become:
$ a_1 = \dfrac{\mathrm{e}^2+\mathrm{e}^n}{\mathrm{e}^{\mathrm{e}^2}}, a_2 = \dfrac{\mathrm{e}^{\mathrm{e}^2+\mathrm{e}^n}}{\mathrm{e}^{\mathrm{e}^{\mathrm{e}^2}}} $, etc, . . .
Does this sequence also converges to zero?
What am I missing?
Thanks everyone
 A: Let's consider the general case, with $a_n=\frac{x_n}{y_n}, x_{n+1}=e^{x_n}, y_{n+1}=e^{y_n}$, where $x_1, y_1>0$ and we wish to study the behaviour as $n \to \infty$.
If $x_1=y_1$, then $x_n=y_n \, \forall n \in \mathbb{Z}^+$, so $a_n=1 \, \forall n \in \mathbb{Z}^+$.
If $x_1<y_1$, then $x_n<y_n \, \forall n \in \mathbb{Z}^+$. We first prove a well known inequality $e^x \geq x+1$ for $x \geq 0$. The function $e^x-x$ attains the value of $1$ at $x=0$, and has gradient $e^x-1 \geq 0$ for $x \geq 0$, so we are done.
Note that $x_{n+1}=e^{x_n}>x_n$, so $x_n \geq x_1$. Thus $y_{n+1}-x_{n+1}=e^{y_n}-e^{x_n}=e^{x_n}(e^{y_n-x_n}-1) \geq e^{x_n}(y_n-x_n) \geq e^{x_1}(y_n-x_n)$. This gives $y_n-x_n \geq (e^{x_1})^{n-1}(y_1-x_1)$, so $0 \leq a_{n+1}=\frac{x_{n+1}}{y_{n+1}}=\frac{1}{e^{y_n-x_n}} \leq \frac{1}{e^{((e^{x_1})^{n-1}(y_1-x_1))}}$. Since $ \frac{1}{e^{((e^{x_1})^{n-1}(y_1-x_1))}}$ converges to $0$, so does $a_n$.
Finally if $x_1>y_1$, then $\frac{1}{a_n}$ converges to $0$ by above, so $a_n$ diverges.
For your first sequence, since $e^2+1<e^{e^2}$, the sequence converges to $0$. For the second, that would depend on the size of $n$. If $e^2+e^n<e^{e^2}$ then the sequence converges to $0$, if $e^2+e^n=e^{e^2}$ then the sequence is always $1$. Else the sequence diverges.
