Inverse Binomial Sum

Calculus Level 2

∑ n = 3 ∞ 1 ( n 3 ) = 1 ( 3 3 ) + 1 ( 4 3 ) + 1 ( 5 3 ) + 1 ( 6 3 ) + ⋯ = ? \displaystyle \sum_{n=3}^{\infty} \dfrac{1}{\binom{n}{3}} = \dfrac{1}{\binom{3}{3}} + \dfrac{1}{\binom{4}{3}} + \dfrac{1}{\binom{5}{3}} + \dfrac{1}{\binom{6}{3}} + \cdots = \, ?


The answer is 1.5.

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Worranat Pakornrat by Worranat Pakornrat , 36, Thailand

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2 solutions

Chew-Seong Cheong
Feb 10, 2017

Relevant wiki: Partial Fractions - Repeated Factors

S = ∑ n = 3 ∞ 1 ( n 3 ) = ∑ n = 3 ∞ 3 × 2 n ( n − 1 ) ( n − 2 ) = ∑ n = 1 ∞ 6 n ( n + 1 ) ( n + 2 ) = 6 2 ∑ n = 1 ∞ ( 1 n − 2 n + 1 + 1 n + 2 ) = 3 ∑ n = 1 ∞ ( 1 n − 1 n + 1 − 1 n + 1 + 1 n + 2 ) = 3 ( 1 − 1 2 ) = 3 2 = 1.5 \begin{aligned} S & = \sum_{n=3}^\infty \frac 1{n \choose 3} \\ & = \sum_{n=3}^\infty \frac {3\times 2}{n(n-1)(n-2)} \\ & = \sum_{\color{#3D99F6}n=1}^\infty \frac {6}{\color{#3D99F6}n(n+1)(n+2)} \\ & = \frac 62 \sum_{n=1}^\infty \left(\frac 1n - \frac 2{n+1} + \frac 1{n+2} \right) \\ & = 3 \sum_{n=1}^\infty \left(\frac 1n - \frac 1{n+1} - \frac 1{n+1} + \frac 1{n+2} \right) \\ & = 3 \left( 1 - \frac 12 \right) = \frac 32 = \boxed{1.5} \end{aligned}

14 Helpful 3 Interesting 6 Brilliant 2 Confused

Sir, i try to solve in the same way but i got halted by 3rd step where i tried breaking them without using partial fraction. Can you tell me any such trick to break without using fraction? :)

Naren Bhandari - 3 years, 8 months ago

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I broke it without using partial fractions:

6 ∑ 1 ( n − 1 ) n ( n − 2 ) 6\sum \dfrac{1}{(n-1)n(n-2)}

3 ∑ 1 n − 1 . n − ( n − 2 ) n ( n − 2 ) 3 \sum \dfrac{1}{n-1} . \dfrac{n - (n - 2)}{n(n-2)}

3 ∑ 1 n − 1 ( 1 n − 2 − 1 n ) 3 \sum \dfrac{1}{n-1} \left( \dfrac{1}{n-2} - \dfrac{1}{n} \right)

I also tried to group n with n- 1 and n - 1 with n - 2 but only this one telescopes.

Shubhrajit Sadhukhan - 4 months, 3 weeks ago

Relevant wiki: Telescoping Series - Sum

∑ n = 3 ∞ 1 ( n 3 ) = 1 ( 3 3 ) + 1 ( 4 3 ) + 1 ( 5 3 ) + 1 ( 6 3 ) + ⋯ = 6 ( 1 1 ⋅ 2 ⋅ 3 + 1 2 ⋅ 3 ⋅ 4 + 1 3 ⋅ 4 ⋅ 5 + 1 4 ⋅ 5 ⋅ 6 + ⋯ ) \displaystyle \sum_{n=3}^{\infty} \dfrac{1}{\binom{n}{3}} = \dfrac{1}{\binom{3}{3}} + \dfrac{1}{\binom{4}{3}} + \dfrac{1}{\binom{5}{3}} + \dfrac{1}{\binom{6}{3}} + \cdots = 6(\dfrac{1}{1\cdot 2\cdot 3} + \dfrac{1}{2\cdot 3\cdot 4} + \dfrac{1}{3\cdot 4\cdot 5} + \dfrac{1}{4\cdot 5\cdot 6} + \cdots)

Observe that 1 1 ⋅ 2 − 1 2 ⋅ 3 = 3 − 1 1 ⋅ 2 ⋅ 3 = 2 1 ⋅ 2 ⋅ 3 \dfrac{1}{1\cdot 2} - \dfrac{1}{2\cdot 3} = \dfrac{3-1}{1\cdot 2\cdot 3} = \dfrac{2}{1\cdot 2\cdot 3} , and it will be similar to other fractions:

6 ( 1 1 ⋅ 2 ⋅ 3 + 1 2 ⋅ 3 ⋅ 4 + 1 3 ⋅ 4 ⋅ 5 + 1 4 ⋅ 5 ⋅ 6 + ⋯ ) = 6 2 ( 1 1 ⋅ 2 − 1 2 ⋅ 3 + 1 2 ⋅ 3 − 1 3 ⋅ 4 + 1 3 ⋅ 4 − 1 4 ⋅ 5 + ⋯ ) 6(\dfrac{1}{1\cdot 2\cdot 3} + \dfrac{1}{2\cdot 3\cdot 4} + \dfrac{1}{3\cdot 4\cdot 5} + \dfrac{1}{4\cdot 5\cdot 6} + \cdots) = \dfrac{6}{2}(\dfrac{1}{1\cdot 2} - \dfrac{1}{2\cdot 3}+\dfrac{1}{2\cdot 3} - \dfrac{1}{3\cdot 4} +\dfrac{1}{3\cdot 4} - \dfrac{1}{4\cdot 5} + \cdots)

All the latter terms are cancelled out, and we are left with:

3 ( 1 2 ) = 1.5 3(\dfrac{1}{2}) = \boxed{1.5}

16 Helpful 1 Interesting 3 Brilliant 1 Confused

Did the same

I Gede Arya Raditya Parameswara - 4 years, 3 months ago

Wrong way bt still got ryt ans

ritik raj - 3 years ago

U checked only 1 term ,what about next term....

ritik raj - 3 years ago

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