Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit Use comparison test to show that $\sum^{+\infty}_{k=1} \frac{1}{k(k+1)(k+2)}$ converges and find its limit

I tried expanding out the denominator, and then using the comparison test with $\frac{1}{k^3}$ but I think this is an incorrect use of the comparison test as I get divergence. I know that the limit is $\frac{1}{4}$ but I am not sure how to use the comparison test to then apply the limit.
$$\begin{align}\sum^\infty_{k=1} \frac{1}{k(k+1)(k+2)} &amp; &lt;\sum^\infty_{k=1} \frac{1}{k(k+0)(k+0)}\\ &amp; =\sum^\infty_{k=1} \frac{1}{k^3}\end{align}$$

It converges.

Franko's telescoping sum:

$$\sum^\infty_{k=1} \frac{1}{k(k+1)(k+2)}=\sum^\infty_{k=1}\left(\frac{1/2}{k(k+1)}-\frac{1/2}{(k+1)(k+2)}\right)$$

$$=\frac14$$
Yes you compare with $\frac{1}{k^3}$, as follows, 
$$\frac{k^3}{k(k+1)(k+2)}\rightarrow 1$$ so the series converges, because $\sum \frac{1}{k^3}$ converges.