Estimate the sum of $\sum _{k=1}^n\left(\frac{1}{\sqrt{k}}\right)$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.