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

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?

• You can use Mark Viola idea here math.stackexchange.com/questions/2149448/… , just reverse the inequality by using $\frac{1}{2\sqrt{k+1}}$ instead in the final argument.
– zwim
Apr 22, 2018 at 17:00
• So $a_1 = -2; a_2 =1-2\sqrt 2; a_3 = 1 + \frac 1{\sqrt 2} - 2\sqrt 3$ etc. If you have $a_{n+1} \le -1 + \frac 1{\sqrt n} \le -1 + \frac 1{\sqrt 2} < 0$ for $n \ge 2$ that's enought to show it is bounded above (by 0!) and if it is monotically increasing it must converge. I don't see anywhere in the question that you must find what it converges to or what the least upper bound is. So you are done. However, if you do have to figure out the value of the limit, you do have more work to do. But DO you have to find the value of the limit? Apr 22, 2018 at 17:35
• Where the "-1" came from? You show by the induction that $a_n<0$ , you cant use -1. Am I missing something? Apr 22, 2018 at 17:49
• "Where the "-1" came from?" Uh.... from you? You said "I'm always stuck with $a_{n+1} \le - 1 + \frac 1{\sqrt n}$ or something like that". If you you got that then that's just fine and you are done. But I don't know how you got $a_{n+1} \le -1 + \frac 1{\sqrt n}$. All I know is that you claimed you did. I'm just saying that if you did. You are actually done. Apr 22, 2018 at 17:57
• Yes, because I tried to prove $a_n<-1$ in that case I used -1.But if you show for $a_n<0$ you cant use what I wrote because it is wrong in that case Apr 22, 2018 at 18:07

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?

• Sorry, but I'm a new BS.c candidate and did not learn integrals yet. I need to use the basic theorems and lemmas from the Calculus course syllabus (what has been taught so far) Apr 22, 2018 at 17:06
• @AlexˢᵅˢʰᵅDruzina Ok, then induction is a direct way. Have you tried it? Apr 22, 2018 at 17:10
• Yes, 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. Apr 22, 2018 at 17:17
• To prove by induction$$\sum_{k=1}^{n-1} \frac{1}{\sqrt{k}}< 2\sqrt{n-1} ,\quad n\ge2,$$ leads to $$2\sqrt{n-1}+\frac1{\sqrt{n}} <2\sqrt{n}.$$ Can you take it from here? Apr 22, 2018 at 17:19
• I'm sorry, but I can't figure out how that would help me :( Can you please elaborate further? Apr 22, 2018 at 17:24

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.

$\frac{1}{\sqrt{x}}$ 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.

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

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