Show the limit of a sequence of real numbers Define $(a_n)_{n=1}^\infty$ as a sequence of real numbers and let a be real number. Show that
$$\lim_{n\rightarrow\infty}(a_n)=a<=>\forall\epsilon\in ]0, 10^{-6}]∃N≥10^{-6}\forall n≥N: |a_n − a|<27\epsilon$$
Where n,N are natural numbers.
Im not sure if Ive understood it correctly. Here is what I have tried: 
if $(a_n)$ tends towards a then $∃\epsilon=\frac{1}{n10^7}\forall n≥10^{-6}$
which means that
$|a_n − a|<\frac{1}{n10^7}\forall n≥10^{-6}$
which means
$|a_n − a|+a_n<a_n<a+\frac{1}{n10^7}$
Taking the limit we get
$\lim_{n\rightarrow\infty}(|a_n − a|+a_n)<\lim_{n\rightarrow\infty}(a_n)<\lim_{n\rightarrow\infty}(a+\frac{1}{n10^7})$
which equals $0+a_n<a_n<a+0$
So we have $\lim_{n\rightarrow\infty}(a_n)=a$

I think it's more like : $\lim\limits_{n\to\infty}(a_n) = a \iff \forall\varepsilon \in \ ]0,10^{-6}]\ \exists N \geq \fbox{$10^6$} \ \forall n\geq N : |a_n - a| < 27\varepsilon$ ?
$\lim\limits_{n\to\infty}(a_n) = a$ is :
$$\forall\varepsilon > 0\ \exists N\ \forall n\geq N : |a_n - a| < \varepsilon$$
is equivalent to 
$$\forall\varepsilon > 0\ \exists N' > 10^6\ \forall n\geq N : |a_n - a| < \varepsilon$$
Because if it's true for a $N < 10^6$, it's also true $\forall N' > 10^6$
Like before, it's true $\forall\varepsilon > 0$ si it's always true for $0 < \varepsilon < 10^{-6}$, and still true for $27\varepsilon$.
So now we have :
$$\lim\limits_{n\to\infty}(a_n) = a \iff \forall\varepsilon\in\ (0,10^{-6}]\ \exists N' > 10^6\ \forall n\geq N : |a_n - a| < 27\varepsilon\ \square$$
