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.

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TopNewest\(\approx0.55086\)

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

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I typed in WolframAlpha and get the result.

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