Let $\displaystyle S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+......+\ frac{1}{\sqrt{80}}$
Find $\displaystyle [S]$ (where [S] denotes the greatest integer less than or equal to S)
Actually I did it like this:
$\displaystyle \sqrt{n+1}>\sqrt{n}$
$\displaystyle \sqrt{n+1}+\sqrt{n}>2\sqrt{n}$
$\displaystyle \frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}$
$\displaystyle \sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}$
Similarly,
$\displaystyle \sqrt{n-1}<\sqrt{n}$
$\displaystyle
\sqrt{n}+\sqrt{n-1}<2\sqrt{n}
$
$\displaystyle \frac{1}{\sqrt{n}+\sqrt{n-1}}>\frac{1}{2\sqrt{n}}$
But how to do with Calculus
$\displaystyle
\sqrt{n}-\sqrt{n-1}>\frac{1}{2\sqrt{n}}
$
for each $\displaystyle k\in\mathbb N$ is $\displaystyle \frac{1}{\sqrt{k+1}}<\int_{k}^{k+1}{\frac{dt}{\sqr t{t}}}<\frac{1}{\sqrt{k}}.$
lower bound: $\displaystyle 2\sqrt{n+1}-2=\int_{1}^{n+1}{\frac{dt}{\sqrt{t}}}=\sum\limits_ {k=1}^{n}{\int_{k}^{k+1}{\frac{dt}{\sqrt{t}}}}<\su m\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}.$
upper bound: $\displaystyle \sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}=1+\sum\l imits_{k=1}^{n-1}{\frac{1}{\sqrt{k+1}}}<1+\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{\frac{dt}{\sqrt{t}}}}=1+2\sqrt{n }-2=2\sqrt{n}-1.$
and we're done.