# Convergence Divergence Assembled!

In this note, I am going to present a list of convergent and divergent series.

• $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k}}$ diverges.

• $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}}$ diverges. You can read the note here.

• $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}}$ converges.

• $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}$ converges. You can read the note here.

• $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)...(k+p)}}$ converges for positive integer $p\geq2$.

First, I will show why $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k}}$ diverges.

The proof is rather straightforward.

We know that $\sqrt{k}\leq k$ for positive integer $k$.

$\Rightarrow \cfrac{1}{\sqrt{k}}\geq \cfrac{1}{k}$

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k}}\geq \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k}$

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k}$ is the harmonic series, it diverges. You can see the proof here.

So, it follows that $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k}}$ diverges.

Next, I am going to show why $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}}$ converges.

I have to admit, the proof is not easy and I have invested a lot of brainpower in it.

$k^2+2k

$\cfrac{1}{k(k+2)}>\cfrac{1}{(k+1)^2}$

$\therefore \cfrac{1}{\sqrt{(k+1)(k+1)(k+1)}}<\cfrac{1}{\sqrt{k(k+1)(k+2)}}<\cfrac{1}{\sqrt{k(k)(k)}}$

$\cfrac{1}{\sqrt{(k+1)^3}}<\cfrac{1}{\sqrt{k(k+1)(k+2)}}<\cfrac{1}{\sqrt{k^3}}$

$\displaystyle{\lim_{n\to \infty}}\cfrac{1}{\sqrt{(k+1)^3}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k^3}}$

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k^3}}=\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{2^3}}+\cfrac{1}{\sqrt{3^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{5^3}}+\cfrac{1}{\sqrt{6^3}}+\cfrac{1}{\sqrt{7^3}}+\cfrac{1}{\sqrt{8^3}}+\cfrac{1}{\sqrt{9^3}}+...$

$\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{9^3}}+\cfrac{1}{\sqrt{9^3}}+\cfrac{1}{\sqrt{9^3}}+\cfrac{1}{\sqrt{9^3}}+\cfrac{1}{\sqrt{9^3}}+...<\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{2^3}}+\cfrac{1}{\sqrt{3^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{5^3}}+\cfrac{1}{\sqrt{6^3}}+\cfrac{1}{\sqrt{7^3}}+\cfrac{1}{\sqrt{8^3}}+\cfrac{1}{\sqrt{9^3}}+...<\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{9^3}}+...$

$\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{9^3}}+\cfrac{1}{\sqrt{9^3}}+\cfrac{1}{\sqrt{9^3}}+\cfrac{1}{\sqrt{9^3}}+\cfrac{1}{\sqrt{9^3}}+...=\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k-1}{\sqrt{(k^2)^3}}=\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k-1}{k^3}=\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} (\cfrac{2}{k^2}-\cfrac{1}{k^3})$

$\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{1^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{4^3}}+\cfrac{1}{\sqrt{9^3}}+...=\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k+1}{\sqrt{(k^2)^3}}=\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k+1}{k^3}=\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} (\cfrac{2}{k^2}+\cfrac{1}{k^3})$

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^2}=1^2+\cfrac{1}{2^2}=\cfrac{1}{3^2}+...=\cfrac{\pi^2}{6}$

So, $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^2}$ converges.

$\cfrac{1}{k^3}<\cfrac{1}{k^2}$

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^3}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^2}=\cfrac{\pi^2}{6}$

It follows that $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^3}$ converges.

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k-1}{k^3}$ and $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k+1}{k^3}$ converges.

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k-1}{k^3}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k^3}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k+1}{k^3}$

$\therefore \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k^3}}$ converges.

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{(k+1)^3}}$ converges.

We have shown that $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{(k+1)^3}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k^3}}$

It follows that $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}}$ converges. The proof is complete.

Whew. That was pretty long.

Now, for the generalization $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)...(k+p)}}$ converges for positive integer $p\geq2$.

$\cfrac{1}{\sqrt{k(k+1)(k+2)}}>\cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}>...>\cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)...(k+p)}}$ for some positive integer $p\geq 2$.

$\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}}>\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)}}>...>\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)...(k+p)}}$ for some positive integer $p\geq 2$.

We have just shown that $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}}$ converges.

$0<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)...(k+p)}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}}$

So it follows that $\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)(k+3)...(k+p)}}$ converges for some positive integer $p\geq 2$. The proof is complete.

That's all for now. It's the longest note that I've ever written.

Do correct me if I'm wrong. Feel free to share your thoughts with me on this note here.

I'm signing off for now. Until next time. Note by Donglin Loo
3 years ago

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You could also start with $k^{p+1} < k(k+1)(k+2)\cdots(k+p) < (k+p)^{p+1}$. Then apply $p$-test.

- 3 years ago

What's a p test

- 3 years ago

This.

- 3 years ago

That was very elaborate and hard to follow at some points but a great proof nonetheless. Honestly you can very quickly see that for $p \geq 3$ the series definitely converges by comparing the series to the Basel Problem Series and showing that the series converges for $p = 3$ and since for $p> 3$ the sum is less than for $p=3$ so it must converge as well. Anyway great problem.

- 2 years, 12 months ago