<s>[INST] Stuck on proof of $\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$ for $n \geq 3$ using induction I'm relatively familiar with induction, I'm just stuck on this step. I am currently taking Introduction to Abstract Math, and have taken Calculus I and II.
$P(n)$ is
$$\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$$
Prove P(n) is true for all $n, \{n \in \mathbb{Z}^+ | \: n \geq 3 \}$
Basis Step
Show that $P(3)$ is true.
$$\sum_{k=1}^{3} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{3}$$
$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2}\leq \frac{7}{4} - \frac{1}{3}$$
$$\frac{49}{36} \leq \frac{17}{12}$$
$\frac{49}{36} \leq \frac{17}{12}$ is true, therefore $P(3)$ is true.
Induction Step
Assume $P(n)$ is true for some $n, \{n \in \mathbb{Z}^+ | \: n \geq 3 \}$. Show that $P(n+1)$ holds true, with $P(n+1)$ being defined as $\sum_{k=1}^{n+1} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n+1}$.
$$\sum_{k=1}^{n} \frac{1}{k^2} \leq \frac{7}{4} - \frac{1}{n}$$
$$\sum_{k=1}^{n} \frac{1}{k^2} + \frac{1}{(n+1)^2} \leq \frac{7}{4} - \frac{1}{n} + \frac{1}{(n+1)^2}$$
That's where I got stuck. What would be the next step?
 [/INST] You know that
$$\sum_{k=1}^n\frac1{k^2}+\frac1{(n+1)^2}\le\frac74-\frac1n+\frac1{(n+1)^2}\,,\tag{1}$$
and you want to show that lefthand side of $(1)$ is at most $\frac74-\frac1{n+1}$; the most straightforward way to do this is to show that
$$\frac74-\frac1n+\frac1{(n+1)^2}\le\frac74-\frac1{n+1}\,,$$
which amounts to showing that
$$\frac1{(n+1)^2}\le\frac1n-\frac1{n+1}\,.$$
And since $\frac1n-\frac1{n+1}=\frac1{n(n+1)}$, this is clear.
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