Question:
If $P(n): \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n}}, n \in N$, then $P(n)$ is true for all $n \geq$ ________________
Solution:
$P(n): \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}} ; n \in N$
for n = 1,
$P(1): 1<\frac{1}{\sqrt{1}}=1$
which is a false statement.
for n = 2,
$P(2): \sqrt{2}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}$
which is true
for n = 3
$P(3): \sqrt{3}<1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}$
which is again true
Hence, $P(n): \sqrt{n}<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\ldots \ldots$ is true for $n \geq 2$.