proof convergence of $(\sqrt[n]{25})_{n\in\mathbb N}$ and $(\frac{2^n}{n!})_{n\in\mathbb N}$ I want to show that $(\sqrt[n]{25})_{n\in\mathbb N}$ and $(\frac{2^n}{n!})_{n\in\mathbb N}$ are convergent.
So for the first one I did the following:\begin{align} 25&=(1+\delta_n)^n \\
\Rightarrow25&\geq1+n\delta_n\\
\Rightarrow \frac{24}{n}&\geq\delta_n\geq0\end{align}
And so $\delta_n\rightarrow0$ and so $\sqrt[n]{25}\rightarrow 1$
But I don't get the second sequence. Are there any tricks? (We've just had the definition of convergence of a sequence and some properties e.g. unique limit) Besides are there any easier steps for the first sequence?
 A: Another answer to the first question: note that $\log\sqrt[n]{25} = \frac1n\log25$ certainly tends to $0$. Since the function $e^x$ is continuous at $x=0$, you can conclude that
$$
\lim_{n\to\infty} \sqrt[n]{25} = \lim_{n\to\infty} e^{\log\sqrt[n]{25}} = e^{\lim_{n\to\infty} (\log\sqrt[n]{25})} = e^0 = 1.
$$
Another answer to the second question: note that
\begin{align*}
0 \le \frac{2^n}{n!} &= \frac{2\cdot2\cdot2\cdots2\cdot2}{1\cdot2\cdot3\cdots(n-1)\cdot n} \\
&= \frac21 \frac22 \frac23 \cdots \frac2{n-1} \frac2n \\
&\le 2\cdot1\cdot1\cdots1\cdot\frac2n = \frac4n.
\end{align*}
Now you can use the squeeze theorem to show that $\frac{2^n}{n!}$ tends to $0$.
A: Well, you can note that $$\cfrac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}=\frac{2^{n+1}n!}{2^n(n+1)!}=\frac2{n+1}$$ for all $n.$ In particular, for $n\ge 2,$ we have $$\cfrac{\frac{2^{n+1}}{(n+1)!}}{\frac{2^n}{n!}}\le\frac23,$$ and so $$\frac{2^{n+1}}{(n+1)!}\le\frac23\cdot\frac{2^n}{n!}.$$ Hence, you have an eventually decreasing sequence of positive numbers, so you can conclude...what?
As for the first one, your approach is just fine.
