Proof by Induction: $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$ I'm having some troubles trying to prove the Exercise 13, page 41 of Apostol's Calculus I, which is the one used to explain some features of integration in the next pages.
It says:
Prove that $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$ if $n\geq 1$. Then use this to prove that 
$$2\sqrt{m}-2 < \sum_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1 \qquad \text{if } m \geq 2$$
(At this moment I'm stuck with the first part of the problem).
Proof (by Induction):
$$P(n): 2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})\qquad  \text{for }n\geq 1$$
Base Case: $P(1)$
$$\begin{align*}
2(\sqrt{1+1} - \sqrt{1})       &<  \frac{1}{\sqrt{1}}  < 2(\sqrt{1}-\sqrt{1-1})&\\ 2(\sqrt{2} - \sqrt{1})         &<\hspace{5pt}  1 \hspace{6pt}    <  2
\end{align*}$$
which is true.
Induction Hypothesis: Assume $P(k)$ is true for a positive integer $k \geq 1$:
$$\underbrace{2(\sqrt{k+1} - \sqrt{k})}_{a} < \frac{1}{\sqrt{k}} < 2(\sqrt{k}-\sqrt{k-1})\qquad (1)$$
Induction Step: Prove $P(k+1)$:
$$2(\sqrt{k+2} - \sqrt{k+1}) < \frac{1}{\sqrt{k+1}} < \underbrace{2(\sqrt{k+1} - \sqrt{k})}_{a}$$

This was my strategy:
Since we assumed $P(k)$ is true, by multiplying the set of inequalities $(1)$ by $\frac{\sqrt{k}}{\sqrt{k+1}}$ we get
$$2\left(\sqrt{k} - \frac{k}{\sqrt{k+1}}\right) < \frac{1}{\sqrt{k+1}} < \underbrace{2\left(\frac{k}{\sqrt{k+1}}-\sqrt{\frac{k-1}{k+1}}\right)}_{b}$$
Then, if we can show that $b < a$, by transitivity we have that $P(k+1)$ is true.
And there is where I'm stuck. I'm not sure what I'm doing wrong, but by setting the inequality $b<a$, it seems that the assumption I made doesn't work:
$$\begin{align*}\frac{k}{\sqrt{k+1}} - \sqrt{\frac{k-1}{k+1}} <&\ \sqrt{k+1} - \sqrt{k}\\ 
\frac{k}{\sqrt{k+1}} - \sqrt{1- \frac{2}{k+1}} <&\ \sqrt{k+1} -
\sqrt{k}\end{align*}$$
The RHS approaches $0$ while the LHS approaches $k$, which reverses the inequality, so I think it's not the correct approach.
Could someone shade me some light on it?
 A: Since $$\sqrt{n+1} - \sqrt{n} = \frac{1}{\sqrt{n+1} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{1}{2\sqrt{n}},$$ we have $$2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}.$$ Since $$\frac{1}{2\sqrt{n}} = \frac{1}{\sqrt{n} + \sqrt{n}} < \frac{1}{\sqrt{n} + \sqrt{n-1}} = \sqrt{n} - \sqrt{n-1}$$ we also have $$\frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1}).$$
A: $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n}-\sqrt{n-1})$ ->multiply with $\sqrt{n}$ everywhere
$2(\sqrt{(n+1)*n} - n) < 1 < 2(n-\sqrt{(n-1)*n})$ ->lets do it in two parts.
$2\sqrt{(n+1)*n} < 2*n+1 $ ->first part starts 
$4*n^2+4*n < 4*n+4*n+1 $ -> take both sides square.(since both is bigger than zero it cannot broke inequality)
$0 < 1 $->tautology
$2*\sqrt{(n-1)*n} < 2*n-1$->second part starts
$4*(n-1)*n < 4*n^2-4*n+1$-> again take square both .
$4*n^2-4*n < 4*n^2-4*n+1$-> open write everything.
$ 0 < 1$-> tautology
It will continue for second part of your question...
$$2\sqrt{m}-2 < \sum_{n=1}^m\frac{1}{\sqrt{n}}< 2\sqrt{m}-1 \qquad \text{if } m \geq 2$$
$ 2*\sqrt{m}+c $  is growing function and $f(k)= 2*\sqrt{k}+c$
$2*\sqrt{m+1}-2*\sqrt{m}$ is the difference between f(m+1) and f(m)
$f_2(m)$ = $\sum_{n=1}^m\frac{1}{\sqrt{n}}$ and $f_2$ is growing function for our domain.
$f_2(m+1)-f_2(m) = \sum_{n=1}^{m+1}\frac{1}{\sqrt{n}}-\sum_{n=1}^m\frac{1}{\sqrt{n}}$
$f_2(m+1)-f_2(m) = \frac{1}{\sqrt{m+1}}$
$\frac{1}{\sqrt{m+1}}-2*\sqrt{m+1}+2*\sqrt{m}->(comparison-op)<-0$
$\frac{1}{\sqrt{m+1}}+2*\sqrt{m}->(comparison-op)<-(2*\sqrt{m+1})$
$1+2*\sqrt{m}*\sqrt{m+1}->(comparison-op)<-(2*m+2)$
$2*\sqrt{m}*\sqrt{m+1}->(comparison-op)<-(2*m+1)$
$4*m*(m+1)->(comparison-op)<-4*m^2+4*m+1$
$4*m^2+4*m->(comparison-op)<-4*m^2+4*m+1$-> comparison op is <
$4*m^2+4*m<4*m^2+4*m+1$
$\frac{1}{\sqrt{m+1}}-2*\sqrt{m+1}+2*\sqrt{m}<0$
$$\sum_{n=1}^m\frac{1}{\sqrt{n}}<2\sqrt{m}-2 <  2\sqrt{m}-1 \qquad \text{if m goes to infinity}$$ 
