In this note, I am going to present a list of convergent and divergent series.
lim n → ∞ ∑ k = 1 n 1 k \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k}} n → ∞ lim k = 1 ∑ n k 1
lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) 1 here .
lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) 1
lim n → ∞ ∑ k = 1 n 1 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)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) ( k + 3 ) 1 here .
lim n → ∞ ∑ k = 1 n 1 k ( 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)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) . . . ( k + p ) 1 p ≥ 2 p\geq2 p ≥ 2
First, I will show why lim n → ∞ ∑ k = 1 n 1 k \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k}} n → ∞ lim k = 1 ∑ n k 1
The proof is rather straightforward.
We know that k ≤ k \sqrt{k}\leq k k ≤ k k k k
⇒ 1 k ≥ 1 k \Rightarrow \cfrac{1}{\sqrt{k}}\geq \cfrac{1}{k} ⇒ k 1 ≥ k 1
lim n → ∞ ∑ k = 1 n 1 k ≥ lim n → ∞ ∑ k = 1 n 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} n → ∞ lim k = 1 ∑ n k 1 ≥ n → ∞ lim k = 1 ∑ n k 1
lim n → ∞ ∑ k = 1 n 1 k \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k} n → ∞ lim k = 1 ∑ n k 1 harmonic series , it diverges. You can see the proof here .
So, it follows that lim n → ∞ ∑ k = 1 n 1 k \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k}} n → ∞ lim k = 1 ∑ n k 1
Next, I am going to show why lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) 1
I have to admit, the proof is not easy and I have invested a lot of brainpower in it.
k 2 + 2 k < k 2 + 2 k + 1 ⇒ k ( k + 2 ) < ( k + 1 ) 2 k^2+2k<k^2+2k+1\Rightarrow k(k+2)<(k+1)^2 k 2 + 2 k < k 2 + 2 k + 1 ⇒ k ( k + 2 ) < ( k + 1 ) 2
1 k ( k + 2 ) > 1 ( k + 1 ) 2 \cfrac{1}{k(k+2)}>\cfrac{1}{(k+1)^2} k ( k + 2 ) 1 > ( k + 1 ) 2 1
∴ 1 ( k + 1 ) ( k + 1 ) ( k + 1 ) < 1 k ( k + 1 ) ( k + 2 ) < 1 k ( 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)}} ∴ ( k + 1 ) ( k + 1 ) ( k + 1 ) 1 < k ( k + 1 ) ( k + 2 ) 1 < k ( k ) ( k ) 1
1 ( k + 1 ) 3 < 1 k ( k + 1 ) ( k + 2 ) < 1 k 3 \cfrac{1}{\sqrt{(k+1)^3}}<\cfrac{1}{\sqrt{k(k+1)(k+2)}}<\cfrac{1}{\sqrt{k^3}} ( k + 1 ) 3 1 < k ( k + 1 ) ( k + 2 ) 1 < k 3 1
lim n → ∞ 1 ( k + 1 ) 3 < lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) < lim n → ∞ ∑ k = 1 n 1 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}} n → ∞ lim ( k + 1 ) 3 1 < n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) 1 < n → ∞ lim k = 1 ∑ n k 3 1
lim n → ∞ ∑ k = 1 n 1 k 3 = 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + 1 6 3 + 1 7 3 + 1 8 3 + 1 9 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}}+... n → ∞ lim k = 1 ∑ n k 3 1 = 1 3 1 + 2 3 1 + 3 3 1 + 4 3 1 + 5 3 1 + 6 3 1 + 7 3 1 + 8 3 1 + 9 3 1 + . . .
1 1 3 + 1 4 3 + 1 4 3 + 1 4 3 + 1 9 3 + 1 9 3 + 1 9 3 + 1 9 3 + 1 9 3 + . . . < 1 1 3 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + 1 6 3 + 1 7 3 + 1 8 3 + 1 9 3 + . . . < 1 1 3 + 1 1 3 + 1 1 3 + 1 4 3 + 1 4 3 + 1 4 3 + 1 4 3 + 1 4 3 + 1 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}}+... 1 3 1 + 4 3 1 + 4 3 1 + 4 3 1 + 9 3 1 + 9 3 1 + 9 3 1 + 9 3 1 + 9 3 1 + . . . < 1 3 1 + 2 3 1 + 3 3 1 + 4 3 1 + 5 3 1 + 6 3 1 + 7 3 1 + 8 3 1 + 9 3 1 + . . . < 1 3 1 + 1 3 1 + 1 3 1 + 4 3 1 + 4 3 1 + 4 3 1 + 4 3 1 + 4 3 1 + 9 3 1 + . . .
1 1 3 + 1 4 3 + 1 4 3 + 1 4 3 + 1 9 3 + 1 9 3 + 1 9 3 + 1 9 3 + 1 9 3 + . . . = lim n → ∞ ∑ k = 1 n 2 k − 1 ( k 2 ) 3 = lim n → ∞ ∑ k = 1 n 2 k − 1 k 3 = lim n → ∞ ∑ k = 1 n ( 2 k 2 − 1 k 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}) 1 3 1 + 4 3 1 + 4 3 1 + 4 3 1 + 9 3 1 + 9 3 1 + 9 3 1 + 9 3 1 + 9 3 1 + . . . = n → ∞ lim k = 1 ∑ n ( k 2 ) 3 2 k − 1 = n → ∞ lim k = 1 ∑ n k 3 2 k − 1 = n → ∞ lim k = 1 ∑ n ( k 2 2 − k 3 1 )
1 1 3 + 1 1 3 + 1 1 3 + 1 4 3 + 1 4 3 + 1 4 3 + 1 4 3 + 1 4 3 + 1 9 3 + . . . = lim n → ∞ ∑ k = 1 n 2 k + 1 ( k 2 ) 3 = lim n → ∞ ∑ k = 1 n 2 k + 1 k 3 = lim n → ∞ ∑ k = 1 n ( 2 k 2 + 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}) 1 3 1 + 1 3 1 + 1 3 1 + 4 3 1 + 4 3 1 + 4 3 1 + 4 3 1 + 4 3 1 + 9 3 1 + . . . = n → ∞ lim k = 1 ∑ n ( k 2 ) 3 2 k + 1 = n → ∞ lim k = 1 ∑ n k 3 2 k + 1 = n → ∞ lim k = 1 ∑ n ( k 2 2 + k 3 1 )
lim n → ∞ ∑ k = 1 n 1 k 2 = 1 2 + 1 2 2 = 1 3 2 + . . . = π 2 6 \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} n → ∞ lim k = 1 ∑ n k 2 1 = 1 2 + 2 2 1 = 3 2 1 + . . . = 6 π 2
So, lim n → ∞ ∑ k = 1 n 1 k 2 \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^2} n → ∞ lim k = 1 ∑ n k 2 1
1 k 3 < 1 k 2 \cfrac{1}{k^3}<\cfrac{1}{k^2} k 3 1 < k 2 1
lim n → ∞ ∑ k = 1 n 1 k 3 < lim n → ∞ ∑ k = 1 n 1 k 2 = π 2 6 \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} n → ∞ lim k = 1 ∑ n k 3 1 < n → ∞ lim k = 1 ∑ n k 2 1 = 6 π 2
It follows that lim n → ∞ ∑ k = 1 n 1 k 3 \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{k^3} n → ∞ lim k = 1 ∑ n k 3 1
lim n → ∞ ∑ k = 1 n 2 k − 1 k 3 \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k-1}{k^3} n → ∞ lim k = 1 ∑ n k 3 2 k − 1 lim n → ∞ ∑ k = 1 n 2 k + 1 k 3 \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k+1}{k^3} n → ∞ lim k = 1 ∑ n k 3 2 k + 1
lim n → ∞ ∑ k = 1 n 2 k − 1 k 3 < lim n → ∞ ∑ k = 1 n 1 k 3 < lim n → ∞ ∑ k = 1 n 2 k + 1 k 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{1}{\sqrt{k^3}}<\displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{2k+1}{k^3} n → ∞ lim k = 1 ∑ n k 3 2 k − 1 < n → ∞ lim k = 1 ∑ n k 3 1 < n → ∞ lim k = 1 ∑ n k 3 2 k + 1
∴ lim n → ∞ ∑ k = 1 n 1 k 3 \therefore \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k^3}} ∴ n → ∞ lim k = 1 ∑ n k 3 1
lim n → ∞ ∑ k = 1 n 1 ( k + 1 ) 3 \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{(k+1)^3}} n → ∞ lim k = 1 ∑ n ( k + 1 ) 3 1
We have shown that lim n → ∞ ∑ k = 1 n 1 ( k + 1 ) 3 < lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) < lim n → ∞ ∑ k = 1 n 1 k 3 \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}} n → ∞ lim k = 1 ∑ n ( k + 1 ) 3 1 < n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) 1 < n → ∞ lim k = 1 ∑ n k 3 1
It follows that lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) 1
Whew. That was pretty long.
Now, for the generalization lim n → ∞ ∑ k = 1 n 1 k ( 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)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) . . . ( k + p ) 1 p ≥ 2 p\geq2 p ≥ 2
1 k ( k + 1 ) ( k + 2 ) > 1 k ( k + 1 ) ( k + 2 ) ( k + 3 ) > . . . > 1 k ( 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)}} k ( k + 1 ) ( k + 2 ) 1 > k ( k + 1 ) ( k + 2 ) ( k + 3 ) 1 > . . . > k ( k + 1 ) ( k + 2 ) ( k + 3 ) . . . ( k + p ) 1 p ≥ 2 p\geq 2 p ≥ 2
lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) > lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) ( k + 3 ) > . . . > lim n → ∞ ∑ k = 1 n 1 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)}}>\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)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) 1 > n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) ( k + 3 ) 1 > . . . > n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) ( k + 3 ) . . . ( k + p ) 1 p ≥ 2 p\geq 2 p ≥ 2
We have just shown that lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) \displaystyle{\lim_{n\to \infty}}\sum_{k=1}^{n} \cfrac{1}{\sqrt{k(k+1)(k+2)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) 1
0 < lim n → ∞ ∑ k = 1 n 1 k ( k + 1 ) ( k + 2 ) ( k + 3 ) . . . ( k + p ) < lim n → ∞ ∑ k = 1 n 1 k ( 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)}} 0 < n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) ( k + 3 ) . . . ( k + p ) 1 < n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) 1
So it follows that lim n → ∞ ∑ k = 1 n 1 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)(k+3)...(k+p)}} n → ∞ lim k = 1 ∑ n k ( k + 1 ) ( k + 2 ) ( k + 3 ) . . . ( k + p ) 1 p ≥ 2 p\geq 2 p ≥ 2
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.
#Calculus
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You could also start with kp+1<k(k+1)(k+2)⋯(k+p)<(k+p)p+1. Then apply p-test.
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What's a p test
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This.
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≥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.