6th March

Calculus Level 3

1 1 ⋅ 2 + 1 + 1 2 2 ⋅ 3 + 1 + 1 2 + 1 3 3 ⋅ 4 + ⋯ = ? \dfrac{1}{1\cdot 2} +\dfrac{1+\frac{1}{2}}{2\cdot 3}+\dfrac{1+\frac{1}{2}+\frac{1}{3}}{3\cdot 4}+\cdots=\, ?


The answer is 1.644.

Dwaipayan Shikari by Dwaipayan Shikari , 17, India

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2 solutions

Above series is ∑ n = 1 ∞ H n n ( n + 1 ) \sum_{n=1}^∞ \dfrac{H_n}{n(n+1)} lim ⁡ N → ∞ ∑ n = 1 N ( H n n − H n n + 1 ) \lim_{ N\rightarrow{∞}}\sum_{n=1}^N (\dfrac{H_n}{n} -\dfrac{H_n}{n+1}) = H 1 1 + ( H 2 2 − H 1 2 ) + ( H 3 3 − H 2 3 ) + ⋯ + H N N − H N + 1 N = \dfrac{H_1}{1}+(\dfrac{H_2}{2}-\dfrac{H_1}{2} )+(\dfrac{H_3}{3}-\dfrac{H_2}{3})+\cdots+\dfrac{H_N}{N}-\dfrac{H_{N+1}}{N} Here H n = ∑ k = 1 n 1 k H_n= \sum_{k=1}^n \dfrac{1}{k} and

H n − H n − 1 = 1 n H_n-H_{n-1}=\dfrac{1}{n} = lim ⁡ N → ∞ 1 + 1 / 2 2 + 1 / 3 3 + 1 / 4 4 + ⋯ + 1 N 2 = \lim_{N\rightarrow{∞}} 1+\dfrac{1/2}{2} +\dfrac{1/3}{3} +\dfrac{1/4}{4} +\cdots+\dfrac{1}{N^2} = 1 + 1 2 2 + 1 3 2 + ⋯ = π 2 6 = 1.644 ⋯ = 1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\cdots= \boxed{\dfrac{π^2}{6}=1.644\cdots}

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The series which is equal to the (pi^2)/6 at the last part is known as the Basel Problem. You can find a good video on it here: youtube.com/watch?v=d-o3eB9sfls

Baha Yesilyurt - 2 months, 3 weeks ago

You first line is not rightly written. As whole series behaves as divergent series. So better find the nth partial sum and rest follows limiting the sum.

Naren Bhandari - 2 months, 3 weeks ago

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Thanks sir ! I will edit my work . Sorry for late reply

Dwaipayan Shikari - 2 months, 2 weeks ago

( R e v i s e d ) \red{\rm (Revised)} The sum can we rewrite as:

S = lim ⁡ n → ∞ ∑ k = 1 n H k k ( k + 1 ) where H k denotes the k th harmonic number. = lim ⁡ n → ∞ ∑ k = 1 n ( H k k − H k k + 1 ) Note that H k + 1 = H k − 1 k + 1 = lim ⁡ n → ∞ ∑ k = 1 n ( H k k − H k + 1 − 1 k + 1 k + 1 ) = lim ⁡ n → ∞ ∑ k = 1 ∞ ( H k k − H k + 1 k + 1 + 1 ( k + 1 ) 2 ) = lim ⁡ n → ∞ ( 1 − H n + 1 n + 1 + ∑ k = 2 n 1 k ) = ∑ k = 1 ∞ 1 k 2 = ζ ( 2 ) = π 2 6 ≈ 1.64 where ζ ( ⋅ ) denotes the Riemann zeta function. \begin{aligned} S & = \lim_{n \to \infty} \sum_{k=1}^n \frac {H_k}{k(k+1)} & \small \blue{\text{where }H_k \text{ denotes the }k\text{th harmonic number.}} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \left(\frac {H_k}k - \frac \blue{H_k}{k+1} \right) & \small \blue{\text{Note that }H_{k+1}=H_k - \frac 1{k+1}} \\ & = \lim_{n \to \infty} \sum_{k=1}^n \left(\frac {H_k}k - \frac \blue{H_{k+1}-\frac 1{k+1}}{k+1} \right) \\ & = \lim_{n \to \infty} \sum_{k=1}^\infty \left(\frac {H_k}k - \frac {H_{k+1}}{k+1} + \frac 1{(k+1)^2} \right) \\ & = \lim_{n \to \infty} \left(1 - \frac {H_{n+1}}{n+1} + \sum_\red{k=2}^n \frac 1k \right) \\ & = \sum_\blue{k=1}^\infty \frac 1{k^2} = \blue \zeta(2) = \frac {\pi^2}6 \approx \boxed{1.64} & \small \blue{\text{where }\zeta(\cdot) \text{ denotes the Riemann zeta function.}} \end{aligned}

References:

0 Helpful 1 Interesting 0 Brilliant 0 Confused

You can't write it this way, since the infinite series ∑ n = 1 ∞ H n n \sum_{n=1}^\infty \frac{H_n}{n} diverges. You have to do so with with a finite sum, and keep track of the error term, so ∑ n = 1 N H n n ( n + 1 ) = ∑ n = 1 N ( H n n − H n n + 1 ) = ∑ n = 1 N ( H n n − H n + 1 n + 1 + 1 ( n + 1 ) 2 ) = ∑ n = 1 N H n n − ∑ n = 2 N + 1 H n n + ∑ n = 2 N + 1 1 n 2 = 1 − H N + 1 N + 1 + ∑ n = 2 N + 1 1 n 2 → ∑ n = 1 ∞ 1 n 2 = 1 6 π 2 \begin{aligned} \sum_{n=1}^N \frac{H_n}{n(n+1)} & = \; \sum_{n=1}^N \left(\frac{H_n}{n} - \frac{H_n}{n+1}\right) \; = \; \sum_{n=1}^N \left(\frac{H_n}{n} - \frac{H_{n+1}}{n+1} + \frac{1}{(n+1)^2}\right) \\ & = \; \sum_{n=1}^N \frac{H_n }{n} - \sum_{n=2}^{N+1}\frac{H_n}{n} + \sum_{n=2}^{N+1} \frac{1}{n^2} \; = \; 1 - \frac{H_{N+1}}{N+1} + \sum_{n=2}^{N+1}\frac{1}{n^2} \\ & \to \; \sum_{n=1}^\infty \frac{1}{n^2} \; = \; \tfrac16\pi^2 \end{aligned} as N → ∞ N \to \infty .

Mark Hennings - 3 months, 1 week ago

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Thanks, I will amend it,

Chew-Seong Cheong - 3 months ago

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