How to show that $a_n=1+1/\sqrt{2}+\cdots+(1/\sqrt{n-1})-2\sqrt{n}$ has an upper bound.

Question

Let $a_n=1+1/\sqrt{2}+\cdots+(1/\sqrt{n-1})-2\sqrt{n}$ while $a_1=-2,n\ge2 $ , I need to prove that $a_n$ converges. I proved that it is monotonically increasing and tried to prove that it is upper-bounded by induction but failed to.

Also, it was told that $a_n$ converges to $-2<L<-1$ so I tried to show by induction that $a_n$ is bounded by $-1$, but I'm always stuck with $a_{n+1} \le -1 + 1/\sqrt{n}$ or something like that. How can I prove that $a_n$ is bounded from above?

Answer 1

Hint. One may use, for $k\ge1$, $$ 2\sqrt{k+1}-2\sqrt{k}= \int_{k}^{k+1}\frac{dx}{\sqrt{x}} < \frac{1}{\sqrt{k}}<\int_{k-1}^{k}\frac{dx}{\sqrt{x}}=2\sqrt{k}-2\sqrt{k-1}. $$ Then, by summing from $k=1$ to $k=n-1$ terms telescope giving $$ 2\sqrt{n}-2<\sum_{k=1}^{n-1} \frac{1}{\sqrt{k}}< 2\sqrt{n-1}, \quad n\ge2. $$ Can you take it from here?

Answer 2

$\frac{1}{\sqrt{x}}$ is a convex function on $\mathbb{R}^+$, hence by the Hermite-Hadamard inequality

$$ 2\sqrt{n}-2=\int_{1}^{n}\frac{dx}{\sqrt{x}}\leq \sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}-\frac{1}{2}+\frac{1}{2\sqrt{n}}\tag{1} $$

$$ 2\sqrt{n-1/2}-\sqrt{2}= \int_{1/2}^{n-1/2}\frac{dx}{\sqrt{x}}\geq \sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}\tag{2}$$ and by rearranging $$ -\left(\frac{3}{2}-\frac{1}{2\sqrt{n}}\right)\leq\left(-2\sqrt{n}+\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}\right) \leq -\left(\sqrt{2}+\frac{1}{2\sqrt{n}}\right).\tag{3}$$ Actually $$ \left(-2\sqrt{n}+\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}\right) \stackrel{\text{CT}}{=} -2+\sum_{k=1}^{n-1}\underbrace{\frac{1}{\sqrt{k}(\sqrt{k}+\sqrt{k+1})^2}}_{\Theta(k^{-3/2})}\stackrel{n\to +\infty}{\longrightarrow}\zeta\left(\tfrac{1}{2}\right)=-1.4603545\ldots\tag{4}$$ where $\text{CT}$ stands for creative telescoping.

Answer 3

Note: To prove something converges you don't have to figure out what it converges to. And to prove something is bounded above you don't have to find the least upper bound; it's enough just to find any upper bound (and prove it is an upper bound).

So

$a_{n+1} - a_n = \frac 1{\sqrt{n}} - 2\sqrt{n+1} + 2\sqrt{n}$.

Claim: $2\sqrt n + \frac 1{\sqrt n} > 2\sqrt{n+1}$ for all natural $n$.

Pf: As all terms are positive and greater than zero...

$2\sqrt n + \frac 1{\sqrt n} > 2\sqrt{n+1} \iff$

$(2\sqrt n + \frac 1{\sqrt n})^2 > (2\sqrt{n+1})^2 \iff $

$4 n + 4 + \frac 1n > 4(n+1) \iff$

$\frac 1n > 0$.

So it it is true.

And so $a_{n+1} - a_n = 2\sqrt{n} + \frac 1{\sqrt n} - 2\sqrt{n+1} > 0$ so $a_{n+1} > a_n$.

So $\{a_i\}$ is monotonically increasing.

Claim: $a_{n} \le -1 + \frac 1{\sqrt{n-1}} \le 0$ for all $n\ge 2$.

Base case: $a_2 = 1 - 2\sqrt{2} \le 1-2 = -1 < -1 + \frac 1{\sqrt{1}} \le 0$.

Inductive step:

If $a_n \le -1 +\frac 1{\sqrt{n-1}}$ then

$a_{n+1} = a_n + \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1}$

$\le -1 +\frac 1{\sqrt{n-1}}+ \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1}$

We proved above that $\frac 1{\sqrt{n-1}}- 2\sqrt{n} + 2\sqrt{n-1} > 0$ so

$a_{n+1} \le -1 +\frac 1{\sqrt{n-1}}+ \frac 1{\sqrt{n}} - 2\sqrt{n} + 2\sqrt{n-1} < -1 +\frac 1{\sqrt{n}} < -1 + 1 = 0$.

So the claim is true.

$a_n \le -1 + \frac 1{\sqrt{n-1}} \le 0$ so $\{a_i\}$ is bounded above by $0$.

Claim 2: $\{a_i\}$ is bounded above by $-1$.

Claim: Let $\epsilon > 0$. Then $\frac 1{\sqrt{n}} < \epsilon \iff n > \frac 1{\epsilon^2}$.

So for all $n + 1 > \frac 1{\epsilon^2}$ then $a_{n+1} \le -1 + \frac 1{\sqrt{n}} < - 1 + \epsilon$. But as $\{a_i\}$ is monotonically increasing. If $m \le n \le \frac 1{\epsilon^2} -1$ then $a_m < a_{n+1} < -1 + \epsilon$ and if $n \ge n+1 > \frac 1{\epsilon^2}$ then $a_m \le -1 + \frac 1{\sqrt{m-1}} < -1 + \epsilon$.

So all $a_i < 1 + \epsilon$ for all $\epsilon > 0$.

So $a_i \le -1$.

And $\{a_i\}$ is monotonically increasing.

So $\{a_i\}$ converges.

That's it. Now, I don't have any idea what it converges to. My proof was crude and ham-fisted. But it was a legitimate proof.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With a Riemann Zeta Function Identity:

\begin{align} &\bbox[#ffd,10px]{\ds{1 + {1 \over \root{2}} + \cdots + {1 \over \root{n - 1}} - 2\root{n}}} = \pars{\sum_{k = 1}^{n}{1 \over \root{k}} - {1 \over \root{n}}} - 2\root{n} \\[5mm] = &\ -\,{1 \over \root{n}} + \pars{\sum_{k = 1}^{n}{1 \over \root{k}} - 2\root{n}} = -\,{1 \over \root{n}} + \bracks{\zeta\pars{1 \over 2} + {1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x} \end{align}

Note that $\ds{0 < {1 \over 2}\int_{n}^{\infty}{\braces{x} \over x^{3/2}}\,\dd x < {1 \over 2}\int_{n}^{\infty}{\dd x \over x^{3/2}} = {1 \over \root{n}} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\Large\to}\,\,\, {\large 0}}$.

such that $$ \bbx{\lim_{n \to \infty}\pars{\bbox[#ffd,10px]{\ds{1 + {1 \over \root{2}} + \cdots + {1 \over \root{n - 1}} - 2\root{n}}}} = \zeta\pars{1 \over 2}} $$