Convergence criteria of sequences of real numbers 
Which of the following conditions implies (imply) the convergence of a sequence $\{x_n\}$ of real numbers?
(A) Given $\varepsilon>0$ there exists an $n_0\in\mathbb N$ such that for all $n\ge n_0,\left|x_{n+1}-x_n\right|<\varepsilon$.
(B) Given $\varepsilon>0$ there exists an $n_0\in\mathbb N$ such that for all $n\ge n_0,\frac1{(x+1)^2}\left|n_{n+1}-x_n\right|<\varepsilon$
(C) Given $\varepsilon>0$ there exists an $n_0\in\mathbb N$ such that for all $n\ge n_0, (n+1)^2\left|x_{n+1}-x_n\right|<\varepsilon$
(D) Given $\varepsilon>0$ there exists an $n_0\in\mathbb N$ such that for all $m,n$ $m\ge n\ge n_0,\left|x_m-x_n\right|<\varepsilon$

Could you help me find correct choices from this question? I think option c is not true but I have trouble regarding rest of the three options.
My doubt: should I solve it by taking particular examples? (which one suits best?) or there are particular theorems regarding it?
 A: A sequence $\{x_n\}$ is converges to $x$ if:

For all $\varepsilon>0$, there exists $n_0 \in \Bbb{N}$ such that $|x-x_n|<\varepsilon$ for all $n \geq n_0$.

It took me a little while to really internalize this definition. I now think of the definition of convergence as "No matter how small of an $\varepsilon$ we choose, we can always go out in the sequence far enough to find an $x_{n_{0}}$ so that the difference between $x$ and any term after $x_{n_{0}}$ will always be less than $\varepsilon$." 
Once you are comfortable with convergence, I would then recommend familiarizing yourself with Cauchy sequences. From here a very useful theorem would be to prove:

Any (real) cauchy sequence is convergent. 

When you prove that result, (or allow yourself to use that result) you will have an easy time proving/disproving $(D)$.

For all $\varepsilon>0$, there exists $N \in \Bbb{N}$ such that $\frac{1}{N} < \varepsilon$.

Once you know this, you will immediately know that $\frac{1}{(N+1)^2}<\frac{1}{N+1}<\frac{1}{N}$ and this may help you deduce the truth or falsity of $(B)$ and $(C)$. It also looks like you have a counterexample for $(A)$.
A: We claim that the first two conditions do not imply the convergence of the sequence, while the latter two do imply convergence.
For A and B, simply take $\forall n: x_n:=\ln n$.
For C and D, let us first introduce a Cauchy sequence:
$\{x_n\}_n$ is Cauchy if $\forall\epsilon',\exists N,\forall n,m\geq N:|x_n-x_m|<\epsilon'$.
There is another practical way to say this (I'll leave it to you to prove that they are equivalent):
$\{x_n\}_n$ is Cauchy if $\forall\epsilon',\exists N,\forall n\geq N,\forall p:|x_{n+p}-x_n|<\epsilon'$.
Next, it's practical to keep in mind that $\mathbb{R}$ is complete.
Now D says precisely that the sequence is Cauchy.
Finally let us see that C implies that the sequence is Cauchy using the alternative statement of Cauchyness. To do this we also need to recall that the $p$-series converges for $p>1$ (see wikipedia). Set $M:=\sum 1/n^2$.
Let $\epsilon'>0$. Pick $\epsilon:=\dfrac{\epsilon'}{M}>0$ in C, and let $n_0$ be the corresponding index. Set $N:=n_0$. Let $n\geq N$ and $p\in\mathbb{N}$. Then we have $|x_{n+p}-x_n|\leq|x_{n+p}-x_{n+p-1}|+...+|x_{n+1}-x_n|<\dfrac{\epsilon}{(n+p)^2}+...+\dfrac{\epsilon}{(n+1)^2} = \dfrac{\epsilon'}{M}\left(\dfrac{1}{(n+p)^2}+...+\dfrac{1}{(n+1)^2}\right)<\epsilon'.$ Observe that the last inequality follows from that $\left(\dfrac{1}{(n+p)^2}+...+\dfrac{1}{(n+1)^2}\right)$ is nothing but a part of the $2$-series (to be precise it is the difference of two partial sums).
