Proof by induction in inequalities

Question

$$\sum^n_{k=1} \frac1{k^3} \le \frac 5 4 - \frac 1 {2n^2}$$ For all $n\ge 2$

Now this really a tough one for me.

The base case holds at $n = 2$

Then i replaced it with $p$ and then $p+1$

I got an inequality where i deduced that $p+1 = p$ and then plus a 1. But its not burging

Answer 1

Assume that the inequality be equal for $n=p$ as you did $$ \sum\limits_{k = 1}^p {{1 \over {k^{\,3} }}} \le {5 \over 4} - {1 \over {2p^{\,2} }} $$

Then it ia also valid for $n=p+1$ $$ \sum\limits_{k = 1}^{p + 1} {{1 \over {k^{\,3} }}} = \sum\limits_{k = 1}^p {{1 \over {k^{\,3} }}} + {1 \over {\left( {p + 1} \right)^{\,3} }} \le \left( {{5 \over 4} - {1 \over {2p^{\,2} }}} \right) + {1 \over {\left( {p + 1} \right)^{\,3} }} \le {5 \over 4} - {1 \over {2\left( {p + 1} \right)^{\,2} }} $$ because in fact, we can write in sequence $$ \eqalign{ & \left( {{5 \over 4} - {1 \over {2p^{\,2} }}} \right) + {1 \over {\left( {p + 1} \right)^{\,3} }} \le {5 \over 4} - {1 \over {2\left( {p + 1} \right)^{\,2} }} \cr & - {1 \over {2p^{\,2} }} + {1 \over {\left( {p + 1} \right)^{\,3} }} \le - {1 \over {2\left( {p + 1} \right)^{\,2} }} \cr & - {{\left( {p + 1} \right)^{\,2} } \over {p^{\,2} }} + {2 \over {\left( {p + 1} \right)}} \le - 1 \cr & - \left( {p + 1} \right)^{\,3} + 2p^{\,2} \le - p^{\,2} \left( {p + 1} \right) \cr & p^{\,2} \left( {p + 1} \right) \le \left( {p + 1} \right)^{\,3} - 2p^{\,2} \cr & p^{\,3} + 3p^{\,2} \le \left( {p + 1} \right)^{\,3} = p^{\,3} + 3p^{\,2} + 3p + 1 \cr} $$

Answer 2

HINT:

Induction hypothesis: Suppose that $\sum^n_{k=1} \frac1{k^3} \le \frac 5 4 - \frac 1 {2n^2}$ holds for some $n$.

Consider $n+1$: $$\sum^{n+1}_{k=1} \frac1{k^3}\le\frac54+\frac1{(n+1)^3}-\frac1{2n^2}$$ and \begin{align}\frac1{(n+1)^3}-\frac1{2n^2}\le\frac1{2(n+1)^2}&\impliedby 2n^2-(n+1)^3\le n^2\\&\impliedby\cdots\end{align}