# 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:$\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}$. 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.

Note by Donglin Loo
1 year, 6 months ago

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

- 1 year, 6 months ago

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.

- 1 year, 6 months ago

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

- 1 year, 6 months ago

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

- 1 year, 6 months ago

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

- 1 year, 6 months ago

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

- 1 year, 6 months ago

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

- 1 year, 6 months ago

@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.

- 1 year, 6 months ago

Good try.

- 1 year, 6 months ago

Don’t really interest many poeple.

- 1 year, 6 months ago

@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.

- 1 year, 6 months ago

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

- 1 year, 6 months ago