Prove that $x_{n}$ is convergent if $|x_{n+1}-x_n|\leq(\frac45)^n$ The question says; prove the sequence $x_{n}$ given by:
$\left | x_{n+1} - x_{n}\right| \leq (\frac{4}{5})^n \ \forall \ n \in \mathbb{N} $
is convergent.
Here is how I approached the problem:
Since the difference between the terms applies to all $n \in \mathbb{N}$, then we know that it is a Cauchy sequence (specifically, a contractile sequence) where we can assume that $\left| x_2 - x_1 \right|=1 $ then $K = \frac{4}{5} <1$
Also since for large values of $n$, $\lim_{n\rightarrow \infty} (\frac{4}{5})^n = 0$ so the constant $(\frac{4}{5})^n$ can replace $\varepsilon$ in the criteria for contractive sequence:
$\left | x_{n+1} - x_{n}\right| \leq \varepsilon $
Then we can have, $\lim_{n\rightarrow \infty} \left|x_{n+1} - x_{n} \right| = 0$
which is a convergent sequence using the contraction principle. 
However, it seems like I should have used another method that uses similar idea but explicitly shows that the difference between $\left|x_{n+1}-x_{n} \right| \leq \frac{4}{5}\left|x_{n}-x_{n-1} \right|$ and so on until I get to $\left | x_{n+1} - x_{n}\right| \leq (\frac{4}{5})^n \left| x_2 - x_1 \right|$. 
I would like to get some feedback and the what are the major flaws in my solution. Suggestions are very welcome
 A: No, what you are saying is not enough. For example, the sequence $$ x_n=1+\frac12+\dots+\frac1n $$ satisfies that $ |x_n-x_{n+1}|=\frac1{n+1}$, which approaches $0$, and yet the sequence diverges. 
(There is also the obvious typo that $\bigl(\frac45\bigr)^n$ is not a constant.)
You are not given nearly enough information to find the limit of the sequence, so really the most direct way (the only way?) of verifying convergence is to argue that the sequence is Cauchy. (I assume you are working on the reals, or at least in some complete metric (normed?) space, so the Cauchy criterion ensures convergence. If this is not the case, one can produce easy counterexamples.)
OK, to check that the sequence is Cauchy, we need to bound $|x_n-x_m|$ (not just $|x_n-x_{n+1}|$), in a way that ensures that $|x_n-x_m|$ can be made arbitrarily small, as long as we restrict $n,m$ to sufficiently large values. 
The obvious attempt (really, again, the only thing we can do) is to use the triangle inequality. For concreteness, suppose $n<m$. We then have that 
 $$ |x_n-x_m|\le |x_n-x_{n+1}|+|x_{n+1}-x_{n+2}|+\dots+|x_{m-1}-x_m|. $$
Using the bounds we are given, this gives us that 
 $$ |x_n-x_m|\le\left(\frac45\right)^n+\left(\frac45\right)^{n+1}+\dots+\left(\frac45\right)^{m-1}. $$
It is here that the specific numbers $\bigl(\frac45\bigr)^n$ become useful. If instead all we had is that $|x_n-x_{n+1}|\to0$, we would not be able to control $|x_n-x_m|$. However, in this case, we have that 
 $$ |x_n-x_m|\le \left(\frac45\right)^n\frac{1-\left(\frac45\right)^{m-n}}{1-\frac45}\le\left(\frac45\right)^n\frac1{1-\frac45}=5\left(\frac45\right)^n. $$
The latter expression can be made as small as desired as long as $n$ is large enough. That is: For any $\epsilon>0$ there is an $N$ such that, provided that $N<n\le m$, we have that $|x_n-x_m|<\epsilon$. This is precisely what it means for the sequence to be Cauchy.

Of course, if you already have established a result ensuring the convergence of a sequence satisfying $|x_{n+1}-x_{n+2}|\le K|x_n-x_{n+1}|$ for all $n$ and some constant $K<1$, then you can readily conclude that the sequence converges. Anyway, if you have such a lemma, most likely its proof was along the lines of the argument I gave. 
A: Hints: Cauchy Sequence and (assuming $\;m\ge n\;$ )
$$|x_m-x_n|=|x_m-x_{m-1}+x_{m-1}-x_{m-2}+x_{m-2}-x_{m-3}+\ldots x_{n+1}-x_n|\le$$
$$\le\sum_{k=0}^{m-n-1}|x_{m-k}-x_{m-k-1}|\le\sum_{k=0}^{m-n-1}\left(\frac45\right)^{m-k-1}$$
