# Math Help - Greatest integer of sum

1. ## Greatest integer of sum

Let $S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+......+\ frac{1}{\sqrt{80}}$

Find $[S]$ (where [S] denotes the greatest integer less than or equal to S)

2. Originally Posted by pankaj
Let $S=1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+......+\ frac{1}{\sqrt{80}}$

Find $[S]$ (where [S] denotes the greatest integer less than or equal to S)
the answer is $16.$ the reason is the following inequality, which can be easily proved by either induction or routine calculus methods:

$\forall \ n \geq 2: \ \ 2 \sqrt{n+1} - 2 < \sum_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n} - 1.$

3. nevermind I bounded it by the wrong integral

4. Use these inequalities:

$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}, \ \forall n\geq 2$

for $n=2, \ 3, \ \ldots,80$

5. Originally Posted by NonCommAlg
the answer is $16.$ the reason is the following inequality, which can be easily proved by either induction or routine calculus methods:

$\forall \ n \geq 2: \ \ 2 \sqrt{n+1} - 2 < \sum_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n} - 1.$

How do we prove by calculus

6. you can actually prove that inequality by using calculus, thus, the rest follows.

7. Actually I did it like this:
$\sqrt{n+1}>\sqrt{n}$

$\sqrt{n+1}+\sqrt{n}>2\sqrt{n}$

$\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}$

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

Similarly,

$\sqrt{n-1}<\sqrt{n}$

$
\sqrt{n}+\sqrt{n-1}<2\sqrt{n}
$

$\frac{1}{\sqrt{n}+\sqrt{n-1}}>\frac{1}{2\sqrt{n}}$

But how to do with Calculus

$
\sqrt{n}-\sqrt{n-1}>\frac{1}{2\sqrt{n}}
$

8. for each $k\in\mathbb N$ is $\frac{1}{\sqrt{k+1}}<\int_{k}^{k+1}{\frac{dt}{\sqr t{t}}}<\frac{1}{\sqrt{k}}.$

lower bound: $2\sqrt{n+1}-2=\int_{1}^{n+1}{\frac{dt}{\sqrt{t}}}=\sum\limits_ {k=1}^{n}{\int_{k}^{k+1}{\frac{dt}{\sqrt{t}}}}<\su m\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}.$

upper bound: $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}}=1+\sum\l imits_{k=1}^{n-1}{\frac{1}{\sqrt{k+1}}}<1+\sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{\frac{dt}{\sqrt{t}}}}=1+2\sqrt{n }-2=2\sqrt{n}-1.$

and we're done.