Estimate the sum of $\sum _{k=1}^n\left(\frac{1}{\sqrt{k}}\right)$ So i am supposed to estimate the sum 
 $\sum _{k=1}^n\left(\frac{1}{\sqrt{k}}\right)$
In the solution for this,
they estimate 

then 

Which i understand is an estimation and then clever multiplication by 1
but then without and further explanation the book states the result as

My question is, how does one come up with the upper and lower bounds with respect to $n$.
I understand perfectly it can easily be proven by induction, but without me seeing the answer, i would not know what to prove in the first place. 
 A: Essentially, the sum becomes
$$\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}}\leq (2\sqrt{1}-2\sqrt{0})+(2\sqrt{2}-2\sqrt{1})+\cdots+(2\sqrt{n}-2\sqrt{n-1})$$
and so most of the terms get cancelled to get $2\sqrt{n}-2\sqrt{0}=2\sqrt{n}$.
This goes similar for the other bound.
A: Yor are dealing with so-called telescope-sums A telescope sum is of the form
$$ \sum_{k=1}^{n} a_{k+1}-a_k = (a_2-a_1)+(a_3-a_2)+...+(a_n-a_{n-1})+(a_{n+1}-a_n) = a_{n+1}-a_1 .$$
In your case, to get an upper bound for your sum, we estimate:
$$\sum_{k=1}^{n} \frac{1}{\sqrt{k}} \leq \sum_{k=1}^{n} 2(\sqrt{k}-\sqrt{k-1})
= 2\sum_{k=1}^{n} (\sqrt{k}-\sqrt{k-1}) =  $$
$$2( (\sqrt{1}-\sqrt{0}) + (\sqrt{2}-\sqrt{1}) + ... +(\sqrt{n-1}-\sqrt{n-2}) + (\sqrt{n}-\sqrt{n-1})) = 2(\sqrt{n}-\sqrt{0}) = 2 \sqrt{n}.$$
Proceed in an analogous way to obtain your lower bound by using telescoping like above.
A: The sum
$$\sum\limits_{k=1}^{n}2 (\sqrt{k+1} -\sqrt{k})$$
 expands into
$$2(\sqrt{2} -1) + 2(\sqrt{3} - \sqrt{2} ) +\cdots  + 2(\sqrt{n+1} - \sqrt{n})$$
terms cancel out ( Telescoping series )
and we are left with 
$$2(\sqrt{n+1} - 1)$$
The same applies to the upper bound.
