<s>[INST] Alternate method to prove this series in a better way Prove that $\frac{1.2 + 2.3 + 3.4 + .....+ n(n + 1)}{n(n + 3)} \ge \frac{n + 1}{4}$ for $n\ge1$
My attempt : 
 Breaking the series into two different series 
$$ S_1 = \sum_{i = 0}^n i^2 = \frac{n(n + 1) (2n + 1)}{6}$$
$$ S_2 = \sum_{i = 0}^n i = \frac{n(n + 1)}{2}$$
Now $$S_1 + S_2 = \frac{n(n + 1)(n + 2)}{3}$$
Writing down LHS and RHS we get :
$$\frac{(n + 1)(n + 2)}{3(n + 3)} \ge \frac{(n + 1)}{4}$$
Multiplying both sides by $$\frac{12}{n + 1}$$ we get : 
$$4(n + 1) \ge 3$$ 
This will always be true $$. ^ . _ . n\ge1$$
Hence proved.
Is there a mistake in my method? Is there a better method to prove this?
 [/INST] We have
$$
\sum_{k=1}^n k (k+1)
= 2 \sum_{k=1}^{n} \binom{k+1}{2}
= 2 \sum_{k=2}^{n+1} \binom{k}{2}
= 2 \binom{n+2}{3}
= \dfrac13 n (n+1) (n+2)
$$
Therefore
$$
\frac{1\cdot 2 + 2\cdot 3 + \cdots + n(n + 1)}{n(n + 3)}
= \frac{ n (n+1) (n+2)}{3n(n + 3)}
= (n+1)\frac{n+2}{3(n + 3)}
\ge (n+1) \frac{1}{4}
$$
because
$$
\frac{n+2}{3(n + 3)} \ge \dfrac14
$$
for all $n\ge 1$.
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