Trying to prove $2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})$ and use this to prove...

Question

I am trying to prove this $2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})$ if $n \ge 1$ and using this to prove $2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( 2\sqrt{m}-1)$ if $m\ge 2$ and in particular i want to show that $m=10^6$ the last inequality is between 1998 and 1999

i am using prove by induction ,first ,for the first statement but i having trouble establish for n=1 and i am confuse how to prove if n is true by hypothesis then n+1.for the second part i didn't get to that part

Edit

reading the answers and comments bellow now i find myself trying to prove the second part that is prove $2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( 2\sqrt{m}-1)$.

My attempt

$2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})\Leftrightarrow \sum^m_{n=1}2( \sqrt{n+1}-\sqrt n )< \sum^m_{n=1}\frac{1}{\sqrt n}<\sum^m_{n=1}2( \sqrt{n}-\sqrt {n-1})\Leftrightarrow $

i assume this to be true

$2\sum^m_{n=1}( \sqrt{n+1}-\sqrt n )< \sum^m_{n=1}\frac{1}{\sqrt n}<2\sum^m_{n=1}( \sqrt{n}-\sqrt {n-1})\Leftrightarrow $

now using telescoping that is $\sum^m_{n=1} (a_n - a_{n-1})=(a_m - a_0)$ i can not make sense of it

Answer 1

Hint: Observe that $$\sqrt{n+1}-\sqrt{n}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}.$$

Answer 2

For a positive decreasing sequence $(a_n)$, we can write the inequality $$\int\limits_{k+1}^{n+1}a(x)dx\leq\sum\limits_{v=k+1}^{n}a_v\leq\int\limits_{k}^na(x)dx$$ Then, in special case, taking $a_n=\frac{1}{\sqrt{n}}$, $a(x)=\frac{1}{\sqrt{x}}$ and $k=n-1$ you get the desired result.

For the other inequality, you can take $k=0$ in the left side of the inequality above . Then take $k=1$ and add the first term $a_1=1$ in the right side.