Harmonic series bounded above by $\sqrt{n}$

Question

Question:

(a) Prove that $$\sqrt{n+1}-\sqrt{n} > \frac{1}{2\sqrt{n+1}}$$

(b) Prove by induction, for $n \geqslant 7$, that $$1 + \frac{1}{2} + \frac13 + \cdots + \frac1n<\sqrt{n}$$

Answer 1

Answer:

(a) Rationalising the numerator, we have:

\begin{align} \sqrt{n+1}-\sqrt{n} &= \frac{(n+1)-n}{\sqrt{n+1}+\sqrt{n}} \\ &=\frac{1}{\sqrt{n+1}+\sqrt{n}} \\ &> \frac{1}{2\sqrt{n+1}} \qquad (\text{since $\sqrt{n}<\sqrt{n+1}$}) \end{align}

(b) The base case is satisfied, since $H_7 = 2.592\ldots < 2.645\ldots = \sqrt{7}$. Assuming true for $n=k$, we will show it to be true for $n=k+1$. Now,

\begin{align} 1 + \frac{1}{2} + \frac13 + \cdots + \frac1k + \frac{1}{k+1} &< \sqrt{k} + \frac{1}{k+1} \qquad (\text{by induction hypothesis})\\ &< \sqrt{k+1} - \frac{1}{2\sqrt{k+1}} + \frac{1}{k+1} \qquad (\text{by part (a)}) \\ &= \sqrt{k+1} - \frac{\sqrt{k+1} - 2}{2(k+1)} \\ &< \sqrt{k+1} \qquad (\text{since $\sqrt{k+1}>2$ for $k \geqslant 7$}) \end{align}

Therefore, by the principle of mathematical induction, the claim is true for all integers $n \geqslant 7$.