Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$ 
Does the following series converge? If it does, determine the appropriate limit.
$$\sum\limits_{k=1}^\infty\left(\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}\right)$$

The only thing i noticed so far is the occurence od the telescopic series via a transformation:
$$\frac{1}{2k}-\frac{1}{k+1}+\frac{1}{2(k+2)}=\frac{1}{k(k+1)(k+2)}$$
The ratio test delivers the result $k/(k+3)$ which renders it unhelpful, so I have to try something else. Now I have been thinking about finding an explicit expression for the partial sums
$$\sum\limits_{k=1}^N\left(\frac{1}{k(k+1)(k+2)}\right)$$
however I neither know, how do so nor do I know whether it suffices to show, that the partial sums will converge for $N\to\infty$ to conclude that the whole series will have a limit.
I need help on how to determine the the explicit expression of the partial sums and would like to know some good suggestions on what to do else.
 A: You’re making it harder than necessary: for each $k\in\Bbb Z^+$ you have
$$\frac1{k(k+1)(k+2)}<\frac1{k^3}\;;$$
Now just use the ordinary comparison test. 
I’m assuming that you’ve already shown that $\sum_{k\ge 1}\frac1{k^p}$ converges for all $p>1$. If not, note that
$$\sum_{k\ge 1}\frac1{k^p}\ge\int_1^\infty\frac{dx}{x^p}\;,$$
which diverges for $p>1$.
Added: In view of the comments, I’ll suggest another approach, a variation on partial fractions. In order to get something that might telescope, you want ideally two terms with denominators that are offset by $1$, so try this decomposition:
$$\frac1{k(k+1)(k+2)}=\frac{A}{k(k+1)}+\frac{B}{(k+1)(k+2)}\;;$$
clearly $B=-A$ and $A=\frac12$, so $$\frac1{k(k+1)(k+2)}=\frac12\left(\frac1{k(k+1)}-\frac1{(k+1)(k+2)}\right)\;,$$ and your partial sum is 
$$\frac12\sum_{k=1}^N\left(\frac1{k(k+1)}-\frac1{(k+1)(k+2)}\right)\;.$$
A: Hint: $\dfrac{1}{k(k+1)(k+2)}=\dfrac{1}{2}\left(\dfrac{1}{k(k+1)}-\dfrac{1}{(k+1)(k+2)}\right).$ Now use the method of differences to find an explicit expression for the partial sums.
A: You can bound each term from above by $k^{-3}$.  Do you know that sum converges?  If not, you can bound that from above by the integral of $x^{-3}$
A: You may write it in the form $x_k=a_{k+2}-2a_{k+1}+a_k$ (you're almost there).  Thus, if you write out the partial sums
$$x_1+ \cdots + \cdots + x_n$$
cancelling out the corresponding intermediate terms, you will see that since $a_n \to 0$, the sum clearly converges.
You're almost there.
