A sequence converges iff the tail converges A sequence converges iff its tail converge. 
This statement is obviously true. But how will one prove this? I appreciate any help!
I already did try to write it out, but the indexing is a lot of trouble. I try to index it right so that I can prove this.

You want to prove that a sequence, say $(a_n)$, converges iff its tail converges. 
$(\rightarrow)$ Suppose $(a_n)$ converges. Let $(a_n)$ converge to $a$. Then, by the definition of the limit point of a function, $\forall \epsilon > 0, \exists N \in \mathbb{N} \textrm{ s.t. } \color{blue}{n \geq N} \implies \left|a_n - a\right| < \epsilon$. Note that this since this implication holds for $n \geq N$, we are basically saying that the tail of $(a_n)$ converges. (Actually, in analysis, we don't care about what happens to the terms outside of the tail; we only care about the terms in the tail, or, more accurately, we only care about the terms as they approach infinity).
$(\leftarrow)$ Suppose the tail of $(a_n)$ converges to, say $b$. This just means that $\forall \epsilon > 0, \exists M \in \mathbb{N} \textrm{ s.t. } m \geq M \implies \left|a_m - b\right| < \epsilon$. Now, from here, you want to show that $(a_n)$ converges. The point is that you have already done that! To show that $(a_n)$ converges, it suffices to find a natural number s.t. all the numbers larger than that natural number, whenever they index $(a_n)$, are such that all the terms of $(a_n)$ lie within a certain $\epsilon$ neighborhood (which we also denote by the difference in absolute value construct, as we have done so far in this proof). You have already done that in showing the existence of $M \in \mathbb{N}$ above.
