prove the given inequality (for series ) For any given $n \in \Bbb N,$ prove that, $$1+{1\over 2^3}+\cdots+{1\over n^3} <{3\over 2}.$$

For $n\ge 3$ the sum is less than
$$1+\frac{1}{8}+\frac{1}{3}\left(\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\cdots+\frac{1}{(n-1)(n)}\right).$$
Note that $\frac{1}{(2)(3)}=\frac{1}{2}-\frac{1}{3}$ and $\frac{1}{(3)(4)}=\frac{1}{3}-\frac{1}{4}$ and $\frac{1}{(4)(5)}=\frac{1}{4}-\frac{1}{5}$ and so on. It follows by cancellation (telescoping) that
$$\frac{1}{(2)(3)}+\frac{1}{(3)(4)}+\frac{1}{(4)(5)}+\cdots+\frac{1}{(n-1)(n)}\lt \frac{1}{2}.$$ 
Thus our full sum is less than $1+\frac{1}{8}+\frac{1}{6}$.
