Question

Show that
1 + 1/√2+. . . +1/√n ≥ √{2(n − 1)} for n∈ N, n> 1
Solve using inequalities

Answer 1

Answer on Question #79263 – Math – Algebra

Question

1. Show that $1 + 1 / \sqrt{2} + \ldots + 1 / \sqrt{n} \geq \sqrt{\{2(n - 1)\}}$ for $n \in \mathbb{N}$, $n > 1$. (Solve using inequalities).

Solution

Consider the right-hand side of the inequality:

$$\sqrt{\{2(n - 1)\}}$$

Fractional part of the number by definition:

$$\begin{array}{l} \{2(n - 1)\} = 2(n - 1) - [2(n - 1)] \\ 0 \leq \{2(n - 1)\} < 1 \\ 0 \leq \{2n - 2\} < 1 \\ \end{array}$$

2 is an integer, so:

$$\begin{array}{l} \{2n - 2\} = \{2n\} \\ 0 \leq \{2n\} < 1 \\ \end{array}$$

By condition $n \in \mathbb{N}$ and $n > 1$. We know that $\mathbb{N} \in \mathbb{Z} \Rightarrow 2n \in \mathbb{Z} \Rightarrow \{2n\} = 0$.

So:

$$\begin{array}{l} \sqrt{\{2(n - 1)\}} = 0 \\ \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}} = \sum_{n=2}^{\infty} \frac{1}{\sqrt{n}} \\ \end{array}$$

When $n = 2$ (it's minimal value of $n$):

$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} > 0$$

Proved.

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