Estimating partial sums $\sum_{n = 1}^m \frac{1}{\sqrt{n}}$

Apostol's Calculus, exercise number I 4.7 13.

Prove that if $n \geq 1$, then $$ 2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1}) $$ and use this to prove that if $m \geq 2$, then $$ 2 \sqrt{m} - 2 < \sum_{n = 1}^m \frac{1}{\sqrt{n}} < 2\sqrt{m} - 1 $$ In particular, when $m = 10^6$ the sum lies between $1998$ and $1999$.

After establishing the trivial case, for $n=1$, I can't think of a way to perform the inductive step:

assuming $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$ and using this, deducing the same for $n+1$ is true as well:

$2(\sqrt{n+2} - \sqrt{n+1}) < \frac{1}{\sqrt{n+1}} < 2(\sqrt{n+1} - \sqrt{n})$

Any help is very appreciated, thanks

Method 1: (the easiest one)

$$\sqrt{x+1}-\sqrt{x}=\dfrac1{\sqrt{x+1}+\sqrt{x}},\tag1$$ so your inequality is equivalent to : $$ \begin{align*} &\dfrac2{\sqrt{n+1}+\sqrt{n}}\lt\dfrac1{\sqrt{n}}\lt\dfrac2{\sqrt{n}+\sqrt{n-1}}\tag2\\&\iff\dfrac1{\sqrt{n+1}+\sqrt{n}}\lt\dfrac1{2\sqrt{n}}\lt\dfrac1{\sqrt{n}+\sqrt{n-1}}\tag3\\&\iff\sqrt{n}+\sqrt{n-1}\lt2\sqrt{n}\lt\sqrt{n+1}+\sqrt{n}\tag4\\&\iff\sqrt{n-1}\lt\sqrt{n}\lt\sqrt{n+1}.\tag5 \end{align*}$$ This is the crucial part, we have : $$ \sqrt{n-1}\lt\color{#C00}{\sqrt{n}\lt\sqrt{n+1}\lt\sqrt{n+2}}.\tag6 $$ Adding to each side $\sqrt{n+1}$ one obtains : $$ \sqrt{n}+\sqrt{n+1}\lt2\sqrt{n+1}\lt\sqrt{n+1}+\sqrt{n+2}.\tag7 $$ Taking the reciprocal of each side and then multiplying by $2$ yields : $$ \dfrac2{\sqrt{n+1}+\sqrt{n+2}}\lt\dfrac1{\sqrt{n+1}}\lt\dfrac2{\sqrt{n}+\sqrt{n+1}}.\tag8 $$

$(2)$ is $P(n)$ and $(8)$ is $P(n+1)$. We showed that $P(n)\implies P(n+1)$. The base case holds. Hence by induction $(2)$ holds for all $n\geqslant1$. $$\tag*{$\tiny\blacksquare$}$$

Method 2:

By multiplying each side of $(4)$ by $\tfrac{\sqrt{n+1}}{\sqrt{n}}$ one gets : $$ \sqrt{n+1}+\dfrac{\sqrt{n+1}\sqrt{n-1}}{\sqrt{n}}\lt2\sqrt{n+1}\lt\sqrt{n+1}+\dfrac{n+1}{\sqrt{n}}.\tag9 $$ Note: This is the same as the $$\dfrac1{\color{red}{\frac{n+1}{\sqrt{n}}}+\sqrt{n+1}}\lt\frac1{2\sqrt{n+1}}\lt‌​\dfrac1{\sqrt{n}+\color{red}{\frac{\sqrt{n+1}\sqrt{n-1}}{\sqrt{n}}}},$$ I referred to in the comment section. Anyway, to prove $(9)$ we first show that: $$\begin{align*} \sqrt{n+1}+\dfrac{\sqrt{n+1}\sqrt{n-1}}{\sqrt{n}}\lt2\sqrt{n+1}&\iff\dfrac{\sqrt{n+1}\sqrt{n-1}}{\sqrt{n}}\lt\sqrt{n+1}\quad\tag{10}\\ &\iff\dfrac{\sqrt{n-1}}{\sqrt{n}}\lt1\tag{11}\\ &\iff\sqrt{n-1}\lt\sqrt{n}.\tag{12} \end{align*}$$ $(12)$ is obviously true hence $(10)$ holds.

Now moving on to showing the second part : $$\begin{align*} 2\sqrt{n+1}\lt\sqrt{n+1}+\dfrac{n+1}{\sqrt{n}}&\iff\sqrt{n+1}\lt\dfrac{n+1}{\sqrt{n}}=\dfrac{\sqrt{n+1}^2}{\sqrt{n}}\tag{13}\\ &\iff1\lt\dfrac{\sqrt{n+1}}{\sqrt{n}}\tag{14}\\ &\iff\sqrt{n}\lt\sqrt{n+1}\tag{15}.\\ \end{align*}$$

$(15)$ is obviously true hence $(13)$ holds. So by this last result and the former we conclude that $(9)$ holds. $$\tag*{$\tiny\blacksquare$}$$