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:k=1n1k(k+1)\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}. I am going to show that the series diverges as nn \to \infty.

For starters, I will rewrite our series as limnk=1n1k(k+1)\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}

With some observation,

we can see that k=1n1k(k+1)<k=1n1k(k)=k=1n1k\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 k=1n1k(k+1)>k=1n1(k+1)(k+1)=k=1n1k+1\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 k=1n1k+1<k=1n1k(k+1)<k=1n1k\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}

limnk=1n1k+1<limnk=1n1k(k+1)<limnk=1n1k\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}

limnk=1n1k+1=12+13+14+...\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k+1}=\cfrac{1}{2}+\cfrac{1}{3}+\cfrac{1}{4}+... and limnk=1n1k=11+12+13+...\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k}=\cfrac{1}{1}+\cfrac{1}{2}+\cfrac{1}{3}+...

limnk=1n1k\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 limnk=1n1k+1\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k+1} diverges.

We conclude that limnk=1n1k(k+1)\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.

Note by Donglin Loo
1 year, 2 months ago

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

X X - 1 year, 2 months ago

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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.

donglin loo - 1 year, 2 months ago

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

X X - 1 year, 2 months ago

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

donglin loo - 1 year, 2 months ago

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@Donglin Loo That calculation is by WolframAlpha.I think it is something about Euler–Mascheroni constant and a series n=0(1n1n2+n)\sum_{n=0}^\infty\left(\frac1n-\frac1{\sqrt{n^2+n}}\right)

X X - 1 year, 2 months ago

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

donglin loo - 1 year, 2 months ago

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

Annie Li - 1 year, 2 months ago

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@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 - 1 year, 2 months ago

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Good try.

Annie Li - 1 year, 2 months ago

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@Annie Li Don’t really interest many poeple.

Annie Li - 1 year, 2 months ago

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@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 - 1 year, 2 months ago

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

Annie Li - 1 year, 2 months ago

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