Harmonic series bounded above by $\sqrt{n}$ 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:
(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 $n \geqslant 7$.
