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.

\(k^2+2k<k^2+2k+1\Rightarrow k(k+2)<(k+1)^2\)

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

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

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

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

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

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

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

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

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

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
2 weeks, 1 day 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.

Pi Han Goh - 1 week, 6 days ago

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What's a p test

Donglin Loo - 1 week, 6 days ago

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This.

Pi Han Goh - 1 week, 6 days ago

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