Convergence Divergence Assembled!

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

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

  • limnk=1n1k(k+1)\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}} diverges. You can read the note here.

  • limnk=1n1k(k+1)(k+2)\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}} converges.

  • limnk=1n1k(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)}} converges. You can read the note here.

  • limnk=1n1k(k+1)(k+2)...(k+p)\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)...(k+p)}} converges for positive integer p2p\geq2.


First, I will show why limnk=1n1k\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k}} diverges.

The proof is rather straightforward.

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

1k1k\Rightarrow \cfrac{1}{\sqrt{k}}\geq \cfrac{1}{k}

limnk=1n1klimnk=1n1k\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}

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


Next, I am going to show why limnk=1n1k(k+1)(k+2)\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.

k2+2k<k2+2k+1k(k+2)<(k+1)2k^2+2k<k^2+2k+1\Rightarrow k(k+2)<(k+1)^2

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

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

1(k+1)3<1k(k+1)(k+2)<1k3\cfrac{1}{\sqrt{(k+1)^3}}<\cfrac{1}{\sqrt{k(k+1)(k+2)}}<\cfrac{1}{\sqrt{k^3}}

limn1(k+1)3<limnk=1n1k(k+1)(k+2)<limnk=1n1k3\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}}

limnk=1n1k3=113+123+133+143+153+163+173+183+193+...\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}}+...

113+143+143+143+193+193+193+193+193+...<113+123+133+143+153+163+173+183+193+...<113+113+113+143+143+143+143+143+193+...\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}}+...

113+143+143+143+193+193+193+193+193+...=limnk=1n2k1(k2)3=limnk=1n2k1k3=limnk=1n(2k21k3)\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})

113+113+113+143+143+143+143+143+193+...=limnk=1n2k+1(k2)3=limnk=1n2k+1k3=limnk=1n(2k2+1k3)\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})

limnk=1n1k2=12+122=132+...=π26\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, limnk=1n1k2\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^2} converges.

1k3<1k2\cfrac{1}{k^3}<\cfrac{1}{k^2}

limnk=1n1k3<limnk=1n1k2=π26\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 limnk=1n1k3\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^3} converges.

limnk=1n2k1k3\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k-1}{k^3} and limnk=1n2k+1k3\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k+1}{k^3} converges.

limnk=1n2k1k3<limnk=1n1k3<limnk=1n2k+1k3\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}

limnk=1n1k3\therefore \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k^3}} converges.

limnk=1n1(k+1)3\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{(k+1)^3}} converges.

We have shown that limnk=1n1(k+1)3<limnk=1n1k(k+1)(k+2)<limnk=1n1k3\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 limnk=1n1k(k+1)(k+2)\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 limnk=1n1k(k+1)(k+2)...(k+p)\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)...(k+p)}} converges for positive integer p2p\geq2.

1k(k+1)(k+2)>1k(k+1)(k+2)(k+3)>...>1k(k+1)(k+2)(k+3)...(k+p)\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 p2p\geq 2.

limnk=1n1k(k+1)(k+2)>limnk=1n1k(k+1)(k+2)(k+3)>...>limnk=1n1k(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)}}>\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 p2p\geq 2.

We have just shown that limnk=1n1k(k+1)(k+2)\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}} converges.

0<limnk=1n1k(k+1)(k+2)(k+3)...(k+p)<limnk=1n1k(k+1)(k+2)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 limnk=1n1k(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)(k+3)...(k+p)}} converges for some positive integer p2p\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
1 year, 1 month ago

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

Pi Han Goh - 1 year, 1 month ago

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What's a p test

donglin loo - 1 year, 1 month ago

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This.

Pi Han Goh - 1 year, 1 month ago

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That was very elaborate and hard to follow at some points but a great proof nonetheless. Honestly you can very quickly see that for p3p \geq 3 the series definitely converges by comparing the series to the Basel Problem Series and showing that the series converges for p=3p = 3 and since for p>3p> 3 the sum is less than for p=3p=3 so it must converge as well. Anyway great problem.

Piero Sarti - 1 year ago

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