# Convergence!

Hey there. Honestly, I can't believe that I am posting two notes on one single day. If you missed out on my previous note, you can find it here.

Today, I would like to share with you an interesting series. It is the evolution of the series that I talked about in my previous note.

To keep things short, the series looks like this: $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}$.

Right now, I am going to show that the series converges.

$\because k^2+3k+2>k^2+3k\Rightarrow (k+1)(k+2)>k(k+3)$

$\Rightarrow \cfrac{1}{(k+1)(k+2)}<\cfrac{1}{k(k+3)}\Rightarrow \cfrac{1}{(k+1)(k+2)(k+1)(k+2)}<\cfrac{1}{k(k+1)(k+2)(k+3)}$

$\cfrac{1}{\sqrt{(k+1)(k+2)(k+1)(k+2)}}<\cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}<\cfrac{1}{\sqrt{k(k+1)k(k+1)}}$.

$\therefore \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{(k+1)(k+2)(k+1)(k+2)}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)k(k+1)}}$.

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{(k+1)(k+2)}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k(k+1)}$.

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{(k+1)(k+2)}=\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} (\cfrac{1}{k+1}-\cfrac{1}{k+2})=\cfrac{1}{2}-\cfrac{1}{3}+\cfrac{1}{3}-\cfrac{1}{4}+\cfrac{1}{4}-\cfrac{1}{5}...=\cfrac{1}{2}$

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

$\therefore \cfrac{1}{2}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}<1$. This is sufficient to show that the series converge.

The thorn in the flesh is that I fail to show the exact value where the series will converge to!

Do share with me your thoughts at the comment section below. If you happen to know how to find the exact value where the series converges, please help.

I am signing off here. Stay tuned for more. Note by Donglin Loo
2 years, 11 months ago

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.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 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"
MathAppears 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

Sort by:

Top Newest

$\approx0.55086$

- 2 years, 11 months ago

Log in to reply

How do you prove that? Can u show me? Thanks

- 2 years, 11 months ago

Log in to reply

I typed in WolframAlpha and get the result.

- 2 years, 11 months ago

Log in to reply

@X X Seriously? Wow. WolframAlpha is so amazing.

- 2 years, 11 months ago

Log in to reply

Donglin Loo, you made it too complicated. By comparison, the k(k)(k)(k) < the radicand. The fourth root of [k(k)(k)(k)] = k^2 < the fourth root of the radicand. So, (1 divided by k^2) > (1 divided by the fourth root of the radicand).

The summation of the left-hand side converges, because the exponent on k is greater than 1.
By comparison, the summation of the right-hand side also converges.

- 2 years, 10 months ago

Log in to reply

@Linda Slovik Are you referring to the one with four terms?

- 2 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...