# Verify proof of $\sum_{k=1}^{n}{k^{-1/2}}<2\sqrt{n}$ for every $n\ge1$

The following is exercise 2.6.20 (c) from Tom Apostol's Calculus Volume 1, I'd like someone to verify my proof. I'm also interested in simpler alternative proofs.

Determine if the following is true and if so prove it: $$\sum_{k=1}^{n}{k^{-1/2}}<2\sqrt{n} \qquad \text{for every} \qquad n\ge1$$

$$2\sqrt{n} = \int_0^nt^{-1/2}dt=\sum_{k=1}^n\int_{k-1}^kt^{-1/2}dt>\sum_{k=1}^n\int_{k-1}^k{k^{-1/2}}dt=\sum_{k=1}^n{k^{-1/2}}$$

• Induction on $n$? – Jyrki Lahtonen Dec 26 '15 at 21:20
• Your proof is good, and is probably the simplest. – vhspdfg Dec 26 '15 at 21:55
• Nice. And can get a lower bound in a similar way. – orangeskid Dec 26 '15 at 22:00

The proof through the integral test is fine. Induction is a viable alternative: let $S_n = \sum_{k=1}^{n}\frac{1}{\sqrt{k}}$.
Since $S_1<2$, it is enough to check that $S_{n+1}-S_n < 2\left(\sqrt{n+1}-\sqrt{n}\right)$, but the last inequality is equivalent to:
$$\frac{1}{\sqrt{n+1}} < \frac{2}{\sqrt{n}+\sqrt{n+1}}$$ that is trivial since $x\to\sqrt{x}$ is an increasing function over $\mathbb{R}^+$.