If the series has a limit of 0, Can I conclude that it is also the sum? Is it possible for a series to have a sum that is equal to $0$ ?
For instance, we are given a series that is,
$\sum_{n=1}^{\infty}$ $\frac{2n+1}{n^2(n+1)^2}$
taking it's limit we have,
$\lim_{n\to\infty}\frac{2n+1}{n^2(n+1)^2}$ = $\lim_{n\to\infty}\frac{\frac{2n}{(n+1)^2}+\frac{1}{(n+1)^2}}{\frac{n^2(n+1)^2}{(n+1)^2}}$ =$\lim_{n\to\infty}\frac{\frac{2n}{(n+1)^2}+ 0}{n^2}$ = $\frac{0}{\infty}$ = $0$
Therefore we can conclude that the series is Convergent, However can I conclude that it's sum is $0$?
 A: No, this is not true. There is the following theorem which states that:
If $(a_n)$ is a sequence with positive terms, and $\sum_{n=1}^{\infty}a_n<+\infty$ (converges to a real number) then $\lim_{n\to \infty}a_n=0.$
So we have : $\sum_{n=1}^{+\infty}a_n<\infty \Longrightarrow \lim_{n\to \infty}a_n=0$ but the converse is not true.
As a counterexample we can set as $a_n=\frac{1}{n}$ which converges to zero, but the series
$$\sum_{n=1}^{\infty}\frac{1}{n}=\infty$$
is not a convergent one (to a real number).
Your question is common misconception by a 1st year calculus student, but be aware of such things in the future, and try to answer them yourself by proving the statement or finding a counterexample! Have a nice day!
A: 
Therefore we can conclude that the series is Convergent, However can I conclude that it's sum is 0?

No, using your approach you cannot conclude that because the inverse of theorem if $\sum a_{n}<+\infty$ then $\lim a_{n}=0$ does not hold. By other hand, all the test only allow you to conclude about convergence and not about sums of series.  You need more careful with that.
We can use the integral's test for to study the covergence.

If $f$ is a continuous, positive and decreasing function, then $$\sum_{n=1}^{+\infty}f(n)<+\infty \iff \int_{1}^{+\infty}f(x)\, {\rm d}x<+\infty$$

Define $f: \begin{cases}\mathbb{R}\longrightarrow \mathbb{R},\\ x\longmapsto \frac{2x+1}{x^{2}(x+1)^{2}}\end{cases}$, then $f$ is continuous,  positive and decreasing for $x\geqslant 1$ (complete of details here). Since $\int_{1}^{+\infty}f(x)\, {\rm d}x=\frac{1}{2}$ the integral converges and therefore $\sum_{n=1}^{+\infty}\frac{2n+1}{n^{2}(n+1)^{2}}$ also converges.
About the sums of series, notice that
$$\sum_{j=1}^{N}\frac{2j+1}{j^{2}(j+1)^{2}}=1-\frac{1}{(N+1)^{2}} \underset{N\to +\infty}{\longrightarrow }1.$$
by telescopic sums and considering the partial fraction expansion
$$\frac{2j+1}{j^{2}(j+1)^{2}}=\frac{1}{j^{2}}-\frac{1}{(j+1)^{2}}.$$
Therefore, $$\sum_{n=1}^{+\infty}\frac{2n+1}{n^{2}(n+1)^{2}}=1.$$
A: No, it is not true. There are plenty of counter examples here, but I choose the simple one.
We know that
$$\log(1+x)=x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+...$$
Now at $x=1$
$$\log(2)=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...$$
which we can write as,
$$\sum_{i=1}^{\infty} \dfrac{(-1)^{n+1}}{n}=\log(2)\neq 0$$
