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.
 A: Hint: A sequence $a_n$ converges to $L$ if for all $\epsilon$ there is $N$ such that for all $n \geq N$, $|a_n-L| < \epsilon$. Now consider the sequence $b_n = a_{n+1}$, which is one particular tail of the original sequence. We claim that it converges to $L$ as well. Given $\epsilon$, we have to find $N$ such that for all $n \geq N$, $|b_n-L| < \epsilon$. Can you think of such an $N$? It might be helpful to consider $N$ which you might get from the sequence $a$ and the same $\epsilon$.
The other direction is similar.
A: Any m-tail of a sequence is when it's 1st m terms are deleted from the original sequence.So the (m+1)th term becomes first, (m+2) the second and so on.Thus the resulting tail is a part of the original sequence.So the convergence of one implies the convergence of the other. That is the motivation behind the proof.Now the actual proof
let (x(m)) be the m tail , then the rth term of the tail is (r+m)th of the original sequence.Similarly the qth term of the original sequence is (q-m)th term of the tail,where q must be >m for the tail to be well defined. Assume that original sequence (x(n)) converges to x. Then for a given epsilon Abs(x(n)-x)k. Thus for the tail it will be (k-m) as the kth term of tail is (k-m) of the original sequence.Thus for the tail Abs(x(m)-x)
A: 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.
