Is this sufficient to show that the partial sums converge? I am trying to show explicitly that the partial sums (for the series $\sum \frac{1}{j(j+1)}$ from j=1 to $\infty$) converge. Would it be sufficient to say that by looking at $\sum \frac{1}{j(j+1)}$ = $\frac{1}{j}-\frac{1}{j+1}$ and $\frac{1}{j}-\frac{1}{j+1} \rightarrow 0$ as $j \rightarrow \infty$? 
There is a theorem in the book that says that if $\sum a_j$ converges, then $a_j \rightarrow 0$ as $j \rightarrow \infty$, but I dont know if this is an iff condition that holds the other way.
 A: Hint. Here the partial sum is easy to handle, you have (a telescoping sum)
$$
\sum_{j=1}^N \frac{1}{j(j+1)}=\sum_{j=1}^N \left(\frac{1}j-\frac{1}{j+1}\right)=1-\frac{1}{N+1},\qquad N\geq1.
$$
This may interest you.
A: You can prove convergence via a comparison test. We have,
$\sum_{j =1}^{j =\infty} \frac{1}{(j+1)j} = \sum_{j =1}^{j =\infty} \frac{1}{j^2 +j}  \leq \sum_{j =1}^{j =\infty} \frac{1}{j^2}$
Now it is known that $\sum_{j =1}^{j =\infty} \frac{1}{j^2}$ converges, and thus 
$\sum_{j =1}^{j =\infty} \frac{1}{(j+1)j}$ converges. 
A: About the theorem, no it is not an if and only if! This theorem is widely used to prove divergence of series from its logical equivalence that states:

Logical Equivalence of the Theorem. If $\mathop {\lim }\limits_{n \to  + \infty } {a_n} \ne 0$  then $\mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{i = p}^n {{a_n}}$ does not exist or diverges.

About the convergence, as your partial sum is a telescoping one you can simply derive a formula for the partial sum and then evaluate the limit of the partial sum (convergence of series)
$$\sum\limits_{j = p}^n {{1 \over {j(j + 1)}}}  = \sum\limits_{j = p}^n {{1 \over j} - } {1 \over {j + 1}} = {1 \over p} - {1 \over {n + 1}}$$
and hence
$$\mathop {\lim }\limits_{n \to  + \infty } \sum\limits_{i = p}^n {{a_n}}  = \mathop {\lim }\limits_{n \to  + \infty } {1 \over p} - {1 \over {n + 1}} = \mathop {\lim }\limits_{n \to  + \infty } {1 \over p} - \mathop {\lim }\limits_{n \to  + \infty } {1 \over {n + 1}} = {1 \over p} - 0 = {1 \over p}$$
