Proving Inequality: \(\frac{1}{n^2}\) Sum < \(\frac{7}{4}\)

[lfdahl]

Prove the inequality:

\[\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} < \frac{7}{4}, \: \:\: \: n\in \mathbb{N}.\]

- without using the well-known result:

\[\lim_{n\rightarrow \infty }\sum_{k=1}^{n}\frac{1}{k^2} = \frac{\pi^2}{6}\]

 

[MarkFL]

My solution:

Spoiler

$$S=\sum_{k=1}^n\left(\frac{1}{k^2}\right)$$

We may write for $3\le n$:

$$S<\frac{5}{4}+\int_2^n x^{-2}\,dx=\frac{7}{4}-\frac{1}{n}$$

Hence, for all $n\in\mathbb{N}$, we find:

$$S<\frac{7}{4}$$

 

[lfdahl]

My solution:

Spoiler

$$S=\sum_{k=1}^n\left(\frac{1}{k^2}\right)$$

We may write for $3\le n$:

$$S<\frac{5}{4}+\int_2^n x^{-2}\,dx=\frac{7}{4}-\frac{1}{n}$$

Hence, for all $n\in\mathbb{N}$, we find:

$$S<\frac{7}{4}$$

Thankyou, MarkFL!, for your participation and for a correct answer!