# 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 $20$?

• Ooops it's 20 not 200 typo sorry – Ron Davidson Jun 15 '13 at 2:12
• You are falling into the trap of panicking and thinking all operations distribute. Even without the square roots, does $\frac11 + \frac12$ have anything to do with $\frac1{1+2}$? – Eric Jablow Jun 15 '13 at 10:58

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 $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{100}} < \int_0^{100}\frac{1}{\sqrt{x}}\,dx =20.$$

• +1 for a really amazing answer. I wouldn't have thought of that method. – Clarinetist Jun 15 '13 at 2:25
• The Riemann sum really is good for more than just annoying first-semester calculus students with sums of powers of integers... – colormegone Jun 15 '13 at 2:28
• @Clarinetist I saw the inequality and immediately thought: hey, people always use integral to estimate the partial sum for a series, why not try on this one. – Shuhao Cao Jun 15 '13 at 2:32
• Note that a sketch makes this method a lot more intuitive! – preferred_anon Jun 15 '13 at 15:54

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 $20$.

• Simply Beautiful. – nbubis Jun 15 '13 at 7:50
• "Straight from the book" :) – rehband Aug 5 '14 at 18:53

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: $$3\cdot 1+ 5\cdot\frac{1}{2}+7\cdot\frac{1}{3}+9\cdot\frac{1}{4}+...= \sum_{n=1}^9\frac{2n+1}{n}=\sum_{n=1}^9 2 +\frac{1}{n} =20.82$$

• Very nice. If we do "exact" for the first $3$ terms, and your estimate for the rest, we get below the target of $20$. – André Nicolas Jun 15 '13 at 2:52
• Beat me to it... :) I thought something similar to the way Oresme demonstrated the divergence of the harmonic series (of course, that's an infinite sum) would be worth a try. – colormegone Jun 15 '13 at 3:20

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,

$\displaystyle 2\sqrt{101}-2<\sum_{k=1}^{100}{\frac{1}{\sqrt{k}}}<2\sqrt{100}-1=19<20.$

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 $$\sum_{i=1}^{100}f(i)\times 1=1+(2)^{-1/2}+\cdots+(100)^{-1/2}$$

$\int^{100}_1 \frac{1}{\sqrt x} = 2\sqrt x|^{100}_1 = 2\times ( \sqrt 100 - \sqrt 1) = 18$

As, $\sum\limits^{100}_{1} \frac{1}{\sqrt x} < \int^{100}_1 \frac{1}{\sqrt x}$,

So, $\sum\limits^{100}_{1} \frac{1}{\sqrt x} < 20$