# Proof by induction: $2(\sqrt n - 1) < \sum\limits_{i=1}^n \left(\frac{1}{\sqrt{i}}\right)$ [duplicate]

Need help with proof by induction for: $$2(\sqrt n - 1) < \sum_{i=1}^n \left(\frac{1}{\sqrt i}\right)$$

For n=1: Good. Assuming for n, trying to proof for (n+1)... Thanks.

## marked as duplicate by mickep, user91500, Servaes, Thomas Andrews, Davide GiraudoOct 29 '15 at 17:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Look carefully at the homework-problem sheet you copied this problem from. Doesn't it say "prove by induction or in any other way"? So you may want to try another way. – uniquesolution Oct 29 '15 at 9:43

## 1 Answer

We can in fact prove a slightly stronger inequality, i.e., $$2\left(\sqrt{n+1}-1\right) < \sum_{i=1}^n \dfrac1{\sqrt{i}}$$ for all $n \in \mathbb{N}$.

At the induction step, we need to prove that $$\dfrac1{\sqrt{n+1}} + 2(\sqrt{n+1}-1) > 2(\sqrt{n+2}-1)$$ i.e., we need to prove that $$\dfrac1{\sqrt{n+1}} > 2(\sqrt{n+2}-\sqrt{n+1})$$

We have $$\dfrac1{\sqrt{n+1}} = \dfrac2{\sqrt{n+1} + \sqrt{n+1}} > \dfrac2{\sqrt{n+1} + \sqrt{n+2}} = 2(\sqrt{n+2}-\sqrt{n+1})$$ which gives us what we want.

Another way is to make use of the fact that $\dfrac1{\sqrt{x}}$ is strictly decreasing and hence $$\dfrac1{\sqrt{i}} > \int_i^{i+1} \dfrac{dx}{\sqrt{x}}$$ Hence, we obtain that $$\sum_{i=1}^n \dfrac1{\sqrt{i}} > \int_1^{n+1} \dfrac{dx}{\sqrt{x}} = 2\left(\sqrt{n+1}-1\right)$$