Proving an inequality with square roots

[Ragnarok7]

This is problem 13 from section I 4.7 of Apostol's Calculus Volume 1:

Prove that \(\displaystyle 2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})\) if \(\displaystyle n\geq 1\). Then use this to prove that \(\displaystyle 2\sqrt{m}-2<\displaystyle\sum_{n=1}^m\frac{1}{\sqrt{n}}<2\sqrt{m}-1\) if \(\displaystyle m\geq 2\).

I have proved the first inequality, and using that I took \(\displaystyle n=1,2,\ldots,m\) to get the following inequalities:

\(\displaystyle 2(\sqrt{2}-\sqrt{1})<\frac{1}{\sqrt{1}}<2(\sqrt{1}-\sqrt{0})\)
\(\displaystyle 2(\sqrt{3}-\sqrt{2})<\frac{1}{\sqrt{2}}<2(\sqrt{2}-\sqrt{1})\)
\(\displaystyle 2(\sqrt{4}-\sqrt{3})<\frac{1}{\sqrt{3}}<2(\sqrt{3}-\sqrt{2})\)
\(\displaystyle \vdots\)
\(\displaystyle 2(\sqrt{m+1}-\sqrt{m})<\frac{1}{\sqrt{m}}<2(\sqrt{m}-\sqrt{m-1})\)

Adding them, I get

\(\displaystyle 2(\sqrt{m+1}-\sqrt{1})<\displaystyle\sum_{n=1}^m\frac{1}{\sqrt{n}}<2\sqrt{m}\)

and, since \(\displaystyle \sqrt{m+1}>\sqrt{m}\), this implies

\(\displaystyle 2\sqrt{m}-2<\displaystyle\sum_{n=1}^m\frac{1}{\sqrt{n}}<2\sqrt{m}\)

so that the left inequality is the desired one. However, I'm unsure how to get the right side. Does anyone have any hints? Thank you!

 

[Opalg]

This is problem 13 from section I 4.7 of Apostol's Calculus Volume 1:

Prove that \(\displaystyle 2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{n-1})\) if \(\displaystyle n\geq 1\). Then use this to prove that \(\displaystyle 2\sqrt{m}-2<\displaystyle\sum_{n=1}^m\frac{1}{\sqrt{n}}<2\sqrt{m}-1\) if \(\displaystyle m\geq 2\).

I have proved the first inequality, and using that I took \(\displaystyle n=1,2,\ldots,m\) to get the following inequalities:

\(\displaystyle 2(\sqrt{2}-\sqrt{1})< \color{red}{\frac{1}{\sqrt{1}}<2(\sqrt{1}-\sqrt{0})}\)
\(\displaystyle 2(\sqrt{3}-\sqrt{2})<\frac{1}{\sqrt{2}}<2(\sqrt{2}-\sqrt{1})\)
\(\displaystyle 2(\sqrt{4}-\sqrt{3})<\frac{1}{\sqrt{3}}<2(\sqrt{3}-\sqrt{2})\)
\(\displaystyle \vdots\)
\(\displaystyle 2(\sqrt{m+1}-\sqrt{m})<\frac{1}{\sqrt{m}}<2(\sqrt{m}-\sqrt{m-1})\)

Adding them, I get

\(\displaystyle 2(\sqrt{m+1}-\sqrt{1})<\displaystyle\sum_{n=1}^m\frac{1}{\sqrt{n}}<2\sqrt{m}\)

and, since \(\displaystyle \sqrt{m+1}>\sqrt{m}\), this implies

\(\displaystyle 2\sqrt{m}-2<\displaystyle\sum_{n=1}^m\frac{1}{\sqrt{n}}<2\sqrt{m}\)

so that the left inequality is the desired one. However, I'm unsure how to get the right side. Does anyone have any hints? Thank you!

The inequality in red, $\frac{1}{\sqrt{1}}<2(\sqrt{1}-\sqrt{0})$, says that $1<2$. You could improve that if you replace it by $\frac{1}{\sqrt{1}}\leqslant 1$. Then when you add the right inequalities, you should get what you want.

 

[Ragnarok7]

Ah! Very nice. Thank you!