Induction on summation inequality stuck on Induction step

Question

working on a fairly simple induction problem, but stuck when decomposing the m+1^st summation and bringing in the I.H to show it holds.

I am bad with formatting so bare with me.

$$\sum_{k=1}^n \frac1{k^2} \leq 2 - \frac1n$$

Base Case: Set $n = 1$ and show that the inequality holds

IH: Assume for all $n = m$, $\sum_{k=1}^m \frac1{k^2} \leq 2 - \frac1m$ holds

I.S:

$$\sum_{k=1}^{m+1} \frac1{k^2} \leq 2 - \frac1{m+1}$$

since we can decompose m+1 to m we get

$$\sum_{k=1}^m \frac1{k^2} + \frac1{m^2} \le 2 - \frac1m$$

then we can use I.H since we have $\sum_{k=1}^m \frac1{k^2} \le 2 - \frac1{m+1}$

so I am trying to show $2 - 1/(m+1) + 1/(m^2)$, but I am stuck here since I cannot manipulate the equation to come to the m+1st equation on the RHS

Answer 1

For the inductive step you have:

$$\sum_{n=1}^{m+1} \frac{1}{k^2} = \sum_{n=1}^{m} \frac{1}{k^2} + \frac{1}{(m+1)^2} \le 2 - \frac{1}{m} + \frac{1}{(m+1)^2} \le 2 - \frac{1}{m+1}$$

The last inequality is true as:

$$\frac{1}{m} - \frac{1}{(m+1)^2} \ge \frac{1}{m+1} \iff (m+1)^2 - m \ge m(m+1) \iff m^2 + m + 1 \ge m^2 + m$$

Answer 2

i think it must be $$\sum_{k=1}^m\frac{1}{k^2}+\frac{1}{(m+1)^2}\le 2-\frac{1}{m+1}$$ and it must be $$\le 2-\frac{1}{m}+ \frac{1}{(m+1)^2}\le 2-\frac{1}{m+1}$$