Question

Show that 1+(1/√2)+...+(1/√n)>= √2 √(n-1) , for n belongs to N , n>1.

Answer 1

Answer on Question #74601 – Math – Discrete Mathematics Question

Show that

$$1 + \frac {1}{\sqrt {2}} + \dots + \frac {1}{\sqrt {n}} \geq \sqrt {2} \sqrt {n - 1}, \text{ for } n \text{ belongs to } N, n > 1$$

Solution

For any $n \geq 2$, let $P_n$ be the statement that

$$1 + \frac {1}{\sqrt {2}} + \dots + \frac {1}{\sqrt {n}} \geq \sqrt {2} \sqrt {n - 1}$$

Base case

The statement $P_2$ says that

$$\begin{array}{l} 1 + \frac {1}{\sqrt {2}} \geq \sqrt {2} \sqrt {2 - 1}, \\ 1 + \sqrt {2} + \frac {1}{2} \geq 2 \\ \sqrt {2} \geq \frac {1}{2} \\ \end{array}$$

which is true.

Inductive case

Fix $k > 2$, and suppose that $P_k$ holds, that is,

$$1 + \frac {1}{\sqrt {2}} + \dots + \frac {1}{\sqrt {k}} \geq \sqrt {2} \sqrt {k - 1}$$

It remains to show that $P_{k+1}$ holds, that is, that

$$\begin{array}{l} 1 + \frac {1}{\sqrt {2}} + \dots + \frac {1}{\sqrt {k}} + \frac {1}{\sqrt {k + 1}} \geq \sqrt {2} \sqrt {k + 1 - 1} \\ \text{By } P_k, 1 + \frac {1}{\sqrt {2}} + \dots + \frac {1}{\sqrt {k}} \geq \sqrt {2} \sqrt {k - 1} \\ 1 + \frac {1}{\sqrt {2}} + \dots + \frac {1}{\sqrt {k}} + \frac {1}{\sqrt {k + 1}} \geq \sqrt {2} \sqrt {k - 1} + \frac {1}{\sqrt {k + 1}} \\ \end{array}$$

Let

$$\sqrt {2} \sqrt {k - 1} + \frac {1}{\sqrt {k + 1}} \geq \sqrt {2} \sqrt {k + 1 - 1}$$

Then

$$2 (k - 1) + 2 \frac {\sqrt {2} \sqrt {k - 1}}{\sqrt {k + 1}} + \frac {1}{k + 1} \geq 2 k$$

Show that

$$\begin{array}{l} 2 \frac {\sqrt {2} \sqrt {k - 1}}{\sqrt {k + 1}} \geq 2 \\ \sqrt {2} \sqrt {k - 1} \geq \sqrt {k + 1} \\ \end{array}$$ $$2k - 2 \geq k + 1$$ $$k \geq 3$$

Hence

$$2 \frac{\sqrt{2} \sqrt{k - 1}}{\sqrt{k + 1}} + \frac{1}{k + 1} > 2 \frac{\sqrt{2} \sqrt{k - 1}}{\sqrt{k + 1}} \geq 2, \text{ for } k \geq 3$$

Therefore, $P_{k+1}$ holds.

Thus by the principle of mathematical induction, for all $n \geq 1$, $P_n$ holds

$$1 + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}} \geq \sqrt{2} \sqrt{n - 1}, \quad n \text{ belongs to } N, n > 1$$

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