Divergence!

I am back again with my second note! This time, I would like to share a divergent series which I find pretty interesting. Oh, just in case you are wondering, I consider my idea original.

Let's not beat around the bush, shall we?

The series is as follows: $\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}$ . I am going to show that the series diverges as $n \to \infty$ .

For starters, I will rewrite our series as $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}$

With some observation,

we can see that $\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}<\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k)}}=\sum_{k=1}^{n} \cfrac{1}{k}$ and $\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}>\sum_{k=1}^{n} \cfrac{1}{\sqrt{(k+1)(k+1)}}=\sum_{k=1}^{n} \cfrac{1}{k+1}$

Combining both, we have $\sum_{k=1}^{n} \cfrac{1}{k+1}<\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}<\sum_{k=1}^{n} \cfrac{1}{k}$

$\Rightarrow \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k+1}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k}$

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k+1}=\cfrac{1}{2}+\cfrac{1}{3}+\cfrac{1}{4}+...$ and $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k}=\cfrac{1}{1}+\cfrac{1}{2}+\cfrac{1}{3}+...$

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k}$ is the harmonic series, it diverges. You can see the proof here. So, it follows that $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k+1}$ diverges.

We conclude that $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}$ diverges. The proof is complete.

That's all for now. Stay tuned for more.

Do correct me if I am wrong. Also, please feel free to share your thoughts at the comment section below.

Markdown

Appears as

*italics* or _italics_

italics

**bold** or __bold__

bold


- bulleted
- list

bulleted
list


1. numbered
2. list

numbered
list

Note: you must add a full line of space before and after lists for them to show up correctly

paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

example link

> This is a quote

This is a quote

    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"

# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"

Math

Appears as

Remember to wrap math in $ ... $ or \[ ... \] to ensure proper formatting.

2 \times 3

2 \times 3

2^{34}

2^{34}

a_{i-1}

a_{i-1}

\frac{2}{3}

\frac{2}{3}

\sqrt{2}

\sqrt{2}

\sum_{i=1}^3

\sum_{i=1}^3

\sin \theta

\sin \theta

\boxed{123}

\boxed{123}

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

What does all this numbers and symbols have to do with the books?

Annie Li - 2 years, 11 months ago

@Annie Li When I used the phrase for the record, I did not mean it can be seen in a book. Instead, I am trying to convey the message that I believe my idea is original as I have not seen this in somewhere else.

donglin loo - 2 years, 11 months ago

Good try.

@Annie Li – Don’t really interest many poeple.

@Annie Li – @Annie Li Sad to know it does not interest you. It's OK. You don't do things to impress everyone. Life is tough if you try to impress everyone.

@Donglin Loo – Life is tough ONLY when you try. You see, i dont care about impression, just give me ice cream, and ill leave.

Is there a limit for $\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{\sqrt{k(k+1)}}-\ln(n)\right)?$

X X - 2 years, 11 months ago

Hmm... I will think about that. It seems like we can do some integration here. Do u know the way to solve that? If so, can you share your solution with me?Thanks.

Now l'm pretty sure that the limit does exists(about 0.019) ,but not sure about the accurate value

@X X – @XX THANKS! I see. Is it calculated by wolfram alpha?

@Donglin Loo – That calculation is by WolframAlpha.I think it is something about Euler–Mascheroni constant and a series $\sum_{n=0}^\infty\left(\frac1n-\frac1{\sqrt{n^2+n}}\right)$

@X X – @X X Havent really heard about that constant before. Thanks for that sharing.

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$