Inequality on reciprocal of squares

Please verify. Is this a correct proof for the following inequality on any $$m \in \mathbb{N}$$: $$\sum_{i=m}^{\infty} \frac{1}{i^2} \leq \frac{2}{2m-1}$$

Proof: Instead, let's consider an equivalent inequality (by absolute convergence it's equivalent): $$\sum_{i=1}^{\infty} \frac{1}{i^2} \leq \sum_{i=1}^{m} \frac{1}{i^2} + \frac{2}{2m+1}$$ Suppose there exists some $$k$$ such that the inequality does not hold. That is, $$\sum_{i=1}^{\infty} \frac{1}{i^2} > \sum_{i=1}^{k} \frac{1}{i^2} + \frac{2}{2k+1}$$ Then, note that for any $$t \in \mathbb{N}$$ $$\frac{1}{(t+1)^2} + \frac{2}{2t+3} - \frac{2}{2t+1} = \frac{1}{(t+1)^2} -\frac{1}{(t+1/2)(t+3/2)}<0 \tag{*}$$ Consider $$A_m = \sum_{i=1}^{m} \frac{1}{i^2} + \frac{2}{2m+1}$$ as a sequence. Then $$(*)$$ shows that this sequence is strictly decreasing. So its limit $$<\sum_{i=1}^{\infty} \frac{1}{i^2}$$. But in reality, its limit is equal to $$\sum_{i=1}^{\infty} \frac{1}{i^2}$$. Contradiction.

Your proof is correct. Alternatively, note that $$i^2>i^2-\frac{1}{4}=\left(i-\frac12\right)\left(i+\frac12\right)$$ for every positive integer $$i$$. Therefore, $$\frac{1}{i^2}<\frac{1}{\left(i-\frac12\right)\left(i+\frac12\right)}=\frac{1}{i-\frac12}-\frac{1}{i+\frac12}$$ for $$i=1,2,3,\ldots$$. This implies $$\sum_{i=m}^\infty\,\frac{1}{i^2}<\sum_{i=m}^\infty\,\left(\frac{1}{i-\frac12}-\frac{1}{i+\frac12}\right)\,,$$ where the right-hand side is a telescopic sum which is equal to $$\frac{1}{m-\frac12}=\frac{2}{2m-1}\,.$$