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: limnk=1n1k(k+1)(k+2)(k+3)\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.

k2+3k+2>k2+3k(k+1)(k+2)>k(k+3)\because k^2+3k+2>k^2+3k\Rightarrow (k+1)(k+2)>k(k+3)

1(k+1)(k+2)<1k(k+3)1(k+1)(k+2)(k+1)(k+2)<1k(k+1)(k+2)(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)}

1(k+1)(k+2)(k+1)(k+2)<1k(k+1)(k+2)(k+3)<1k(k+1)k(k+1)\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)}}.

limnk=1n1(k+1)(k+2)(k+1)(k+2)<limnk=1n1k(k+1)(k+2)(k+3)<limnk=1n1k(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)}}.

limnk=1n1(k+1)(k+2)<limnk=1n1k(k+1)(k+2)(k+3)<limnk=1n1k(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)}.

limnk=1n1(k+1)(k+2)=limnk=1n(1k+11k+2)=1213+1314+1415...=12\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}

limnk=1n1k(k+1)=limnk=1n(1k1k+1)=1112+1213+1314+...=1\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

12<limnk=1n1k(k+1)(k+2)(k+3)<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
1 year, 5 months ago

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0.55086\approx0.55086

X X - 1 year, 5 months ago

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

donglin loo - 1 year, 5 months ago

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I typed in WolframAlpha and get the result.

X X - 1 year, 5 months ago

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

donglin loo - 1 year, 5 months ago

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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.

Linda Slovik - 1 year, 4 months ago

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

donglin loo - 1 year, 4 months ago

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