Squared fractional sum inequality

Question

Let $n$ be a positive integer. Is it true that $$\sum_{k=n+1}^\infty\dfrac{1}{k^2}<\dfrac1n?$$

Answer 1

Yes, it is true. This is because $$k^2 > k(k-1) \implies \dfrac1{k^2} < \dfrac1{k(k-1)} = \dfrac1{k-1} - \dfrac1k$$ Hence, $$\sum_{k=n+1}^{\infty} \dfrac1{k^2} < \sum_{k=n+1}^{\infty} \left(\dfrac1{k-1} - \dfrac1k \right) = \dfrac1n$$

Answer 2

Yes. We can show this by noting that $f(t) = 1/t^2$ is a decreasing function, and that its definite integrals are therefore underestimated by the corresponding right-sided Riemann sum. Thus, we have $$ \sum_{k=n+1}^ \infty \frac 1{k^2} < \sum_{k=n+1}^ \infty \int_{k-1}^k \frac 1{t^2}dt = \int_n^\infty \frac 1{t^2}dt=\frac 1n $$