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.

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## Comments

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

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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.

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

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

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

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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.

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Good try.

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*with ice-creams.Log in to reply

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