Proving sequence is bounded below?

Question

I have the function: $$ a_{n}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots +\frac{1}{\sqrt{n}}-2\sqrt{n} $$

How could I go about proving that this sequence is bounded from below? When plotting it, I can see it converges to around -1.45, but I can't figure out how to try and prove it's bounded arithmetically. I was trying to figure out how to manipulate it to use the arithmetic/geometric mean inequality, but had no such luck.

Thanks!

Answer 1

hint: $\dfrac{1}{\sqrt{k}}= \dfrac{2}{2\sqrt{k}} \ge \dfrac{2}{\sqrt{k+1}+\sqrt{k}}= 2\sqrt{k+1}- 2\sqrt{k}$, and take the sum for $k$ runs from $1$ to $n-1$.

Answer 2

If you look at the graph of $y=\frac1{\sqrt{x}}$, it’s clear that

$$\sum_{k=1}^n\frac1{\sqrt{k}}\ge\int_1^{n+1}x^{-1/2}\,dx=2\left[\sqrt{x}\right]_1^{n+1}=2\sqrt{n+1}-2$$

and hence that

$$a_n\ge 2\sqrt{n+1}-2-2\sqrt{n}=2\left(\sqrt{n+1}-\sqrt{n}\right)-2>-2\;.$$

Answer 3

You may notice that $$ a_{n+1}-a_{n} = \frac{1}{\sqrt{n+1}}+2\sqrt{n}-2\sqrt{n+1}=\frac{1}{\sqrt{n+1}}-\frac{1}{\frac{1}{2}(\sqrt{n}+\sqrt{n+1})}<0 $$ hence the sequence $\{a_n\}_{n\geq 1}$ is decreasing and $$a_N-a_1 = \sum_{n=1}^{N-1}(a_{n+1}-a_n) = -\sum_{n=1}^{N-1}\frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n+1})^2} $$ so that $$ a_N \geq a_1-\sum_{n=1}^{+\infty}\frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n+1})^2} $$ where the last series is clearly convergent, since its general term behaves like $\frac{1}{n^{3/2}}$.
You may also check that $$ \frac{1}{\sqrt{n}(\sqrt{n}+\sqrt{n+1})^2}\leq \frac{1}{2\sqrt{n-\frac{1}{6}}}-\frac{1}{2\sqrt{n+\frac{5}{6}}} $$ from which $$ a_N \geq \color{red}{-1-\sqrt{\frac{3}{10}}} $$ follows by creative telescoping.

Answer 4

If $a_{n}=\sum_{k=1}^n \frac1{\sqrt{k}}-2\sqrt{n} $, then

$\begin{array}\\ d(n) &=a_{n+1}-a_n\\ &=\dfrac1{\sqrt{n+1}}-2(\sqrt{n+1}-\sqrt{n})\\ &=\dfrac1{\sqrt{n+1}}-2(\sqrt{n+1}-\sqrt{n})\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}\\ &=\dfrac1{\sqrt{n+1}}-2\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\\ &<\dfrac1{\sqrt{n+1}}-2\dfrac{1}{2\sqrt{n+1}}\\ &=0\\ \text{and}\\ d(n) &>\dfrac1{\sqrt{n+1}}-2\dfrac{1}{2\sqrt{n}}\\ &=\dfrac1{\sqrt{n+1}}-\dfrac{1}{\sqrt{n}}\\ &=\dfrac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}\sqrt{n+1}}\\ &=\dfrac{\sqrt{n}-\sqrt{n+1}}{\sqrt{n}\sqrt{n+1}}\dfrac{\sqrt{n}+\sqrt{n+1}}{\sqrt{n}+\sqrt{n+1}}\\ &=\dfrac{-1}{\sqrt{n}\sqrt{n+1}(\sqrt{n}+\sqrt{n+1})}\\ &>\dfrac{-1}{2n\sqrt{n}}\\ \text{so}\\ a_{n+k}-a_n &=\sum_{j=1}^k (a(n+j)-a(n+j-1))\\ &< 0\\ \text{and}\\ a_{n+k}-a_n &=\sum_{j=1}^k (a(n+j)-a(n+j-1))\\ &>\sum_{j=1}^k \dfrac{-1}{2(n+j-1)\sqrt{n+j-1}}\\ \end{array} $

Therefore $a_n$ is a decreasing sequence which is bounded below since $\sum \dfrac1{n^{3/2}} $ converges.

For an explicit bound, since $\dfrac1{x^{3/2}} $ is decreasing for $x \ge 1$, $\dfrac1{x^{3/2}} \lt \int_{x-1}^x \dfrac{dt}{t^{3/2}} $ so

$\begin{array}\\ \sum_{j=1}^k \dfrac{1}{(n+j-1)\sqrt{n+j-1}} &\lt\sum_{j=1}^k \int_{n+j-2}^{n+j-1} \dfrac{dt}{t^{3/2}}\\ &\lt\int_{n-1}^{n+k-1} \dfrac{dt}{t^{3/2}}\\ &\lt\int_{n-1}^{\infty} \dfrac{dt}{t^{3/2}}\\ &\lt\dfrac{2}{\sqrt{n-1}}\\ \end{array} $

so

$\begin{array}\\ a_{n+k}-a_n &>\sum_{j=1}^k \dfrac{-1}{2(n+j-1)\sqrt{n+j-1}}\\ &\gt\dfrac{-1}{\sqrt{n-1}}\\ \text{or}\\ a_{n+k} &>a_n-\dfrac{1}{\sqrt{n-1}}\\ \end{array} $

As is often the case, nothing here is original and I worked it out as I went along.