How to obtain this upper bound on the summation from this inequality?

Question

I can show that $$ \frac{1}{\sqrt{n}} < 2 (\sqrt{n} - \sqrt{n-1} ) $$ for $n \geq 1$. Now from this how to derive the following inequality? $$ \sum_{n=1}^m \frac{1}{\sqrt{n}} < 2\sqrt{m} - 1 $$ for all $m \geq 1$.

By taking $n = 1, 2, \ldots, m$ in the first inequality and then adding the resulting inequalities, we obtain $$ \sum_{n=1}^m \frac{1}{\sqrt{n}} < 2 \sum_{n=1}^m (\sqrt{n} - \sqrt{n-1} ) = 2 \sqrt{m}. $$

But how to derive the inequality I'm looking for? That is, how to derive $$ \sum_{n=1}^m \frac{1}{\sqrt{n}} < 2\sqrt{m} - 1 $$ for all $m \geq 1$ from $$ \frac{1}{\sqrt{n}} < 2 (\sqrt{n} - \sqrt{n-1} ) $$ for $n \geq 1$?

Answer 1

Instead of applying the formula to all the terms, you can start from the second one to get:

\begin{align} \sum_{n=1}^m\frac{1}{\sqrt{n}}=&1+\sum_{n=2}^{m}\frac{1}{\sqrt n}\\ <&1+2\sum_{n=2}^m \left(\sqrt n-\sqrt{n-1}\right)\\ =&1+2\sqrt m-2\\ =&2\sqrt{m}-1 \end{align}

Answer 2

Note that $\sqrt{n}-\sqrt{n-1}=\frac{1}{\sqrt{n}+\sqrt{n-1}}$. (We rationalized the numerator, multiplying top and (missing) bottom by $\sqrt{n}+\sqrt{n-1}$.)

Since $\sqrt{n-1}\lt \sqrt{n}$, we have $$\frac{1}{\sqrt{n}+\sqrt{n-1}}\gt \frac{1}{2\sqrt{n}}.$$