How can I show that the following is true?
$$\sum_{k=1}^{\infty}{\frac{1}{k\sqrt{k}}} < 3$$
I just need this series to be between 1 and 3 so that I can conclude some result. I have found the appropriate lower bound series $$\sum_{k=1}^{\infty}{\frac{1}{k^2}} =\frac{\pi^2}{6}\leq\sum_{k=1}^{\infty}{\frac{1}{k\sqrt{k}}},$$ but am struggling to find an upper one. Could somebody lend me a hand, please?
On
Drew a new one. There is an incorrect $a-1$ on the lower left.
If we have $f(x) > 0$ but $f'(x) < 0,$ then $$ \int_a^{b+1} \; f(x) \; dx \; < \; \sum_{k=a}^b \; f(k) \; < \; \int_{a-1}^b \; f(x) \; dx $$
This has $f(x) > 0$ and $f'(x) < 0.$ The drawing shows an upper bound for the sum.
On
I thought it might be instructive to present an approach that does not require knowledge of integrals. To that end we proceed.
First we can evaluate the telescoping series
$$ \sum_{k=1}^\infty \left(k^{-1/2}-(k+1)^{-1/2}\right)=1\tag1$$
Next, we can rewrite $(1)$ as
$$ \sum_{k=1}^\infty \frac1{k^{1/2}(k+1)^{1/2}\left(k^{1/2}+(k+1)^{1/2}\right)}=1\tag 2$$
Clearly we have from $(2)$
$$\sum_{k=2}^\infty \frac1{2k^{3/2}} < 1$$
from which we arrive at the coveted inequality
$$\sum_{k=1}^\infty \frac1{k^{3/2}}<3$$
And we are done.
On
Working along the same lines as Mark Viola's answer, we can get upper and lower bounds.
Combining fractions, we get $$ \begin{align} \frac2{\sqrt{k}}-\frac2{\sqrt{k+1}} &=2\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}\tag{1a}\\ &=\frac2{\sqrt{k(k+1)}\left(\sqrt{k+1}+\sqrt{k}\right)}\tag{1b} \end{align} $$ Replacing $k$ with $k+1$ in $\text{(1b)}$, we get a lower bound. Replacing $k+1$ with $k$ in $\text{(1b)}$, we get an upper bound. $$ \frac1{(k+1)\sqrt{k+1}}\le\frac2{\sqrt{k}}-\frac2{\sqrt{k+1}}\le\frac1{k\sqrt{k}}\tag2 $$ Summing in $k$ gives $$ \sum_{k=2}^\infty\frac1{k\sqrt{k}}\le2\le\sum_{k=1}^\infty\frac1{k\sqrt{k}}\tag3 $$ Adding $1$ to the left inequality yields $$ 2\le\sum_{k=1}^\infty\frac1{k\sqrt{k}}\le3\tag4 $$
Motivation
The motivation for using $\frac2{\sqrt{k}}$ is because $$ \int_k^\infty\frac{\mathrm{d}x}{x\sqrt{x}}=\frac2{\sqrt{k}}\tag5 $$ Thus, $$ \int_k^{k+1}\frac{\mathrm{d}x}{x\sqrt{x}}=\frac2{\sqrt{k}}-\frac2{\sqrt{k+1}}\tag6 $$ and from $(6)$, we see that $$ \frac1{(k+1)\sqrt{k+1}}\le\int_k^{k+1}\frac{\mathrm{d}x}{x\sqrt{x}}\le\frac1{k\sqrt{k}}\tag7 $$ Putting together $(6)$ and $(7)$, we get $(2)$.
A variant of the approaches presented so far is given by the Hermite-Hadamard inequality. $\frac{1}{x\sqrt{x}}$ is a convex function on $\mathbb{R}^+$, hence
$$ I(n)=2-\frac{2}{\sqrt{n}}=\int_{1}^{n}\frac{dx}{x\sqrt{x}} $$ fulfills $$ I(n)\leq \frac{1}{2}\left(1+\frac{1}{n\sqrt{n}}\right)+\sum_{k=2}^{n-1}\frac{1}{k\sqrt{k}} $$ such that $$ \sum_{k=1}^{n}\frac{1}{k\sqrt{k}} \geq \frac{5}{2}\left(1-\frac{1}{\sqrt{n}}\right). \tag{LowerBound}$$ On the other hand if we consider $$ J(n) = \int_{3/2}^{n+1/2}\frac{dx}{x\sqrt{x}}= \sqrt{\frac{8}{3}}-\frac{4}{\sqrt{4n+2}}$$ we have $$ J(n) \geq \sum_{k=2}^{n}\frac{1}{k\sqrt{k}} $$ hence $$ \sum_{k=1}^{n}\frac{1}{k\sqrt{k}}\leq \color{red}{1+\sqrt{\frac{8}{3}}} < 3.\tag{UpperBound} $$ This bound is pretty sharp: the actual difference between $\zeta\left(\frac{3}{2}\right)$ and $1+\sqrt{\frac{8}{3}}$ is just a bit greater than $0.02$.