Prove Inequality Challenge for $a\in \Bbb{Z^+}$

[anemone]

Let $a\in \Bbb{Z^+}$, prove that $\dfrac{2}{2-\sqrt{2}}>\dfrac{1}{1\sqrt{1}}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+\cdots+\dfrac{1}{a\sqrt{a}}$.

 

[MarkFL]

My solution:

Spoiler

Let:

\(\displaystyle S\equiv\sum_{k=1}^{\infty}\left(k^{-\frac{3}{2}}\right)\)

Using graphical tools, we can easily see that:

\(\displaystyle S<1+\int_{1}^{\infty}u^{-\frac{3}{2}}\,du\)

To evaluate the integral $I$, we may write:

\(\displaystyle I=\lim_{t\to\infty}\left(\int_{1}^{t}u^{-\frac{3}{2}}\,du\right)\)

Applying the anti-derivative form of the FTOC, there results:

\(\displaystyle I=-2\lim_{t\to\infty}\left(\left.u^{-\frac{1}{2}}\right|_{1}^t\right)=2\)

Hence, we have $S<3$ and we may state

\(\displaystyle S_a<S<3<\frac{2}{2-\sqrt{2}}=2+\sqrt{2}\)

Shown as desired. :)

 

[anemone]

My solution:

Spoiler

Let:

\(\displaystyle S\equiv\sum_{k=1}^{\infty}\left(k^{-\frac{3}{2}}\right)\)

Using graphical tools, we can easily see that:

\(\displaystyle S<1+\int_{1}^{\infty}u^{-\frac{3}{2}}\,du\)

To evaluate the integral $I$, we may write:

\(\displaystyle I=\lim_{t\to\infty}\left(\int_{1}^{t}u^{-\frac{3}{2}}\,du\right)\)

Applying the anti-derivative form of the FTOC, there results:

\(\displaystyle I=-2\lim_{t\to\infty}\left(\left.u^{-\frac{1}{2}}\right|_{1}^t\right)=2\)

Hence, we have $S<3$ and we may state

\(\displaystyle S_a<S<3<\frac{2}{2-\sqrt{2}}=2+\sqrt{2}\)

Shown as desired. :)

Awesome, MarkFL!(Yes) And I welcome others to try it with other approach too!(Happy)

 

[anemone]

Hint to solve the challenge using elementary method:

Spoiler

Telescoping series...

 

[Albert1]

Let $a\in \Bbb{Z^+}$, prove that $\dfrac{2}{2-\sqrt{2}}>\dfrac{1}{1\sqrt{1}}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+\cdots+\dfrac{1}{a\sqrt{a}}$.

my solution:

Spoiler

$A=\dfrac{2}{2-\sqrt{2}}=\dfrac {1}{1-\dfrac{\sqrt 2}{2}}$
$B=\dfrac{1}{1\sqrt{1}}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+\cdots+\dfrac{1}{a\sqrt{a}}$
$ =1+\dfrac {\sqrt 2}{2^2}+\dfrac {\sqrt 3}{3^2}+\dfrac {\sqrt 4}{4^2}+------+\dfrac {\sqrt a}{a^2}$
$A$ is a infinite geometric sum with ratio $r=\dfrac {\sqrt 2}{2}$, and first term $A_1=1$
$\therefore A>B$

 

[kaliprasad]

my solution:

Spoiler

$A=\dfrac{2}{2-\sqrt{2}}=\dfrac {1}{1-\dfrac{\sqrt 2}{2}}$
$B=\dfrac{1}{1\sqrt{1}}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+\cdots+\dfrac{1}{a\sqrt{a}}$
$ =1+\dfrac {\sqrt 2}{2^2}+\dfrac {\sqrt 3}{3^2}+\dfrac {\sqrt 4}{4^2}+------+\dfrac {\sqrt a}{a^2}$
$A$ is a infinite geometric sum with ratio $r=\dfrac {\sqrt 2}{2}$, and first term $A_1=1$
$\therefore A>B$

not correct because

Spoiler

nth term of A = $\dfrac{1}{2^{\frac{n}{2}}}$
and nth term of B =$\dfrac{1}{n\sqrt n}$
after certain terms the term of A is smaller

 

[Albert1]

Spoiler

$A=\dfrac{2}{2-\sqrt{2}}=\dfrac {1}{1-\dfrac{\sqrt 2}{2}}=2+\sqrt 2$
$B=\dfrac{1}{1\sqrt{1}}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+\cdots+\dfrac{1}{a\sqrt{a}}$
$A$ is a infinite geometric sum with ratio $r=\dfrac {\sqrt 2}{2}$, and first term $a_1=1 , a_n=\dfrac {1}{2^{(n-1)/2}}$
$B$ is a $P$ series with p=1.5 ,$b_1=1$ ,and $b_n=\dfrac {1}{n^{1.5}}$ Both $A$ and $B$ convergent
$B=\sum\dfrac{1}{n^{1.5}}=\dfrac{1}{1\sqrt{1}}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+\cdots+\dfrac{1}{a\sqrt{a}}$
$B<1+1+\dfrac {1}{2}+\dfrac {1}{4}+\dfrac{1}{8}+------=1+2=3<A$
I did not use "Telescoping series"
hope anemone can show us how to use "Telescoping series"

 

[anemone]

Solution of other:

Spoiler

Note that it's always true that for all $a\in \Bbb{Z^+}$, $\dfrac{1}{4a^3}<\dfrac{1}{4a^3-3a-1}$.

From here we get

$\dfrac{1}{4a^3}<\dfrac{1}{(a-1)(2a+1)^2}$

$\dfrac{(2a+1)^2}{a}<\dfrac{4a^2}{a-1}$

$\dfrac{2a+1}{\sqrt{a}}<\dfrac{2a}{\sqrt{a-1}}$

$\dfrac{2a+1}{a\sqrt{a}}<\dfrac{2}{\sqrt{a-1}}$

$\dfrac{1}{a\sqrt{a}}<\dfrac{2}{\sqrt{a-1}}-\dfrac{2}{\sqrt{a}}$

So we see

\(\displaystyle \begin{align*}\sum_{i=2}^{a}\dfrac{1}{a\sqrt{a}}&<\dfrac{2}{\sqrt{1}}-\dfrac{2}{\sqrt{2}}+\dfrac{2}{\sqrt{2}}-\dfrac{2}{\sqrt{3}}+\dfrac{2}{\sqrt{3}}-\dfrac{2}{\sqrt{4}}+\cdots+\dfrac{2}{\sqrt{a-2}}-\dfrac{2}{\sqrt{a-1}}+\dfrac{2}{\sqrt{a-1}}-\dfrac{2}{\sqrt{a}}\\&<\dfrac{2}{\sqrt{1}}-\dfrac{2}{\sqrt{a}}\end{align*}\)

and therefore we get

$\begin{align*}\dfrac{1}{1\sqrt{1}}+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+\cdots+\dfrac{1}{a\sqrt{a}}&<1+\dfrac{2}{\sqrt{1}}-\dfrac{2}{\sqrt{a}}\\&<3-\dfrac{2}{\sqrt{a}}<3+\dfrac{1}{\sqrt{2}+1}=\dfrac{2}{2-\sqrt{2}}\,\,\,\text{(Q.E.D)}\end{align*}$