Proving that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$ How am I suppose to prove that:
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$$
Do I use the way like how we count $1+2+ \cdots+100$ to estimate?
So $1/5050 \lt 20$, implying that it is indeed less than $20$?
 A: We can estimate the sum by integration: because the inverse square root function is strictly descreasing, hence for $n\in \mathbb{N}$
$$
\frac{1}{\sqrt{n}}< \frac{1}{\sqrt{x}} \quad \text{ for } x\in (n-1,n).
$$
Integrating both sides on this interval:
$$
\frac{1}{\sqrt{n}}= \int^n_{n-1}\frac{1}{\sqrt{n}}\,dx< \int^{n}_{n-1}\frac{1}{\sqrt{x}}\,dx.
$$
Therefore
$$
\frac{1}{\sqrt{1}} < \int_0^1 \frac{1}{\sqrt{x}}\,dx
\\
\frac{1}{\sqrt{2}} < \int_1^2 \frac{1}{\sqrt{x}}\,dx
\\
\cdots
\\
\frac{1}{\sqrt{100}} < \int_{99}^{100} \frac{1}{\sqrt{x}}\,dx
$$
and
$$
\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{100}} < \int_0^{100}\frac{1}{\sqrt{x}}\,dx =20.
$$
A: Start from $\sqrt{n}-\sqrt{n-1}$. Multiply top and "bottom" by $\sqrt{n}+\sqrt{n-1}$. 
We get $\dfrac{1}{\sqrt{n}+\sqrt{n-1}}$, which is $\gt \dfrac{1}{2\sqrt{n}}$. It follows that $\dfrac{1}{\sqrt{n}}\lt 2(\sqrt{n}-\sqrt{n-1})$.
Add up, $n=1$ to $n=100$. We get 
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots +\frac{1}{\sqrt{100}}\lt 2(\sqrt{1}-\sqrt{0})+2(\sqrt{2}-\sqrt{1})+\cdots +2(\sqrt{100}-\sqrt{99}).$$
Observe the mass cancellation on the right: it collapses to $20$.  
A: Mean Value Theorem can also be used,
Let $\displaystyle f(x)=\sqrt{x}$
$\displaystyle f'(x)=\frac{1}{2}\frac{1}{\sqrt{x}}$
Using mean value theorem we have:
$\displaystyle \frac{f(n+1)-f(n)}{(n+1)-n}=f'(c)$ for  some $c\in(n,n+1)$
$\displaystyle \Rightarrow \frac{\sqrt{n+1}-\sqrt{n}}{1}=\frac{1}{2}\frac{1}{\sqrt{c}}$....(1)
$\displaystyle \frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{c}}<\frac{1}{\sqrt{n}}$
Using the above ineq. in $(1)$ we have,
$\displaystyle \frac{1}{2\sqrt{n+1}}<\sqrt{n+1}-\sqrt{n}<\frac{1}{2\sqrt{n}}$
Adding the left part of the inequality we have,$\displaystyle\sum_{k=2}^{n}\frac{1}{2\sqrt{k}}<\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=\sqrt{n}-1$
$\Rightarrow \displaystyle\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2(\sqrt{n}-1)$
$\Rightarrow \displaystyle1+\sum_{k=2}^{n}\frac{1}{\sqrt{k}}<1+2\sum_{k=2}^{n}(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}-2+1=2\sqrt{n}-1$
$\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}<2\sqrt{n}-1$
Similarly adding the right side of the inequality we have,
$\displaystyle\sum_{k=1}^{n}\frac{1}{2\sqrt{k}}>\sum_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})=\sqrt{n+1}-1$
$\Rightarrow \displaystyle\sum_{k=1}^{n}\frac{1}{\sqrt{k}}>2(\sqrt{n+1}-1)$
Showing that,
$\displaystyle 2\sqrt{n+1}-2<\sum_{k=1}^{n}{\frac{1}{\sqrt{k}}}<2\sqrt{n}-1.\tag 1$
Put $n=100$ to prove,
$\displaystyle 2\sqrt{101}-2<\sum_{k=1}^{100}{\frac{1}{\sqrt{k}}}<2\sqrt{100}-1=19<20.$
A: Note that the area under the function $f(x)=x^{-1/2}$ is bigger than the sum  of the retangles of bases 1 and high $f(i)$  where $i=1,\cdots,100$. so you have:
$$\sum_{i=1}^{100}f(i)\times 1\leq \int_{0}^{100}x^{-1/2}dx= 2(100)^{1/2}=20$$
and note that $$\sum_{i=1}^{100}f(i)\times 1=1+(2)^{-1/2}+\cdots+(100)^{-1/2}$$
A: A close value can be obtained easily.
Let's look at the squares until $100$:
$1, 4, 9, 16, 25, 36, 49, 64, 81$.
We then know that the sum will be smallar than:
$$3\cdot 1+ 5\cdot\frac{1}{2}+7\cdot\frac{1}{3}+9\cdot\frac{1}{4}+...=
\sum_{n=1}^9\frac{2n+1}{n}=\sum_{n=1}^9 2 +\frac{1}{n} =20.82$$
A: $\int^{100}_1 \frac{1}{\sqrt x} = 2\sqrt x|^{100}_1 = 2\times ( \sqrt 100 - \sqrt 1) = 18$
As, $\sum\limits^{100}_{1} \frac{1}{\sqrt x} < \int^{100}_1 \frac{1}{\sqrt x}$,
So, $\sum\limits^{100}_{1} \frac{1}{\sqrt x} < 20$
