Serious Series

Calculus Level 5

T ( n ) = 0.75 n 2 1 − 2 2 + 3 3 + 4 − 5 2 + 6 3 + ⋯ (3n terms) \large\mathfrak{T}(n)=\dfrac{0.75n^2}{1-2^2+3^3+4-5^2+6^3+\cdots \text{(3n terms)}}

S ( n ) = 1 T ( n ) + 13 n + 9 9 \large\mathfrak{S}(n)=\dfrac{\frac{1}{\mathfrak{T}(n)}+13n+9}{9}

I = ∑ n = 1 ∞ ( 1 S ( n ) ) = π α β \large\mathfrak{I}=\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{\mathfrak{S}(n)}\right)=\dfrac{\pi^{\alpha}}{\beta}

S ( α ! + β ) = ? \large\mathfrak{S}(\alpha!+\beta)=\, ?

Details and Assumptions

  • n , α , β ∈ Z n,\alpha,\beta\in\mathbb Z and n ≥ 1 n\geq 1


The answer is 20.

Rishabh Jain by Rishabh Jain , 22, Germany

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2 solutions

Rishabh Jain
Apr 14, 2016

Denominator of T n : \mathfrak{T}_n: ( 1 + 4 + ⋯ + ( 3 n − 2 ) ) − ( 2 2 + 5 2 + ⋯ + ( 3 n − 1 ) 2 ) + 3 3 ( 1 3 + 2 3 + ⋯ + n 3 ) \color{#D61F06}{(1+4+\cdots+(3n-2))}\\-\color{#3D99F6}{(2^2+5^2+\cdots+(3n-1)^2)}\\+3^3\color{#20A900}{(1^3+2^3+\cdots+n^3)}

= n 2 ( 2 + ( n − 1 ) 3 ) + ( ∑ r = 1 n ( 3 n − 1 ) 2 ) ⏟ 9 ∑ r = 1 n r 2 − 6 ∑ r = 1 n r + ∑ r = 1 n 1 + 27 ( ( n 2 ) ( n + 1 ) 2 4 ) =\color{#D61F06}{\frac n2(2+(n-1)3)}+\underbrace{\color{#3D99F6}{\left(\displaystyle\sum_{r=1}^{n}\left(3n-1\right)^2\right)}}_{ 9\displaystyle\sum_{r=1}^{n}r^2-6\displaystyle\sum_{r=1}^{n}r+\displaystyle\sum_{r=1}^{n}1}+27\color{#20A900}{\left(\dfrac{(n^2)(n+1)^2}{4}\right)}

= 3 n 2 ( 9 n 2 + 14 n + 9 ) 4 =\dfrac{3n^2(9n^2+14n+9)}{4}

T ( n ) = 1 9 n 2 + 14 n + 9 \large\mathfrak{T}(n)=\dfrac{1}{9n^2+14n+9} S ( n ) = n 2 + 3 n + 2 \large \mathfrak S(n)=n^2+3n+2

I = ∑ n = 1 ∞ ( 1 n 2 + 3 n + 2 ) = ∑ n = 1 ∞ ( 1 n + 1 − 1 n + 2 ) \large\mathfrak{I}=\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n^2+3n+2}\right) \\=\displaystyle\sum_{n=1}^{\infty}\left(\dfrac{1}{n+1}-\dfrac{1}{n+2}\right)

( A T e l e s c o p i c S e r i e s ) \mathbf{(A Telescopic Series)}

= 1 2 = π 0 2 \large =\dfrac{1}{2}=\dfrac{\pi^0}{2}

∴ S ( 3 ) = 20 \therefore\huge\mathfrak{S}(3)=\boxed{20}

10 Helpful 0 Interesting 0 Brilliant 0 Confused

I was trying to find a reasonable appx. for 1/2 ( pi^2/20 , pi^4/195) . Damn !

Keshav Tiwari - 5 years ago

Great !!perfectly designed question and ans too... will u try this https://brilliant.org/problems/how-fast-u-can-approach/

Suneel Kumar - 5 years, 2 months ago

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Yeah it had been posted many times on brilliant and as usual beautiful brilliant minds provided different beautiful solutions... See here ...

Rishabh Jain - 5 years, 2 months ago
Aakash Khandelwal
Apr 14, 2016

When one finds α = 0 \alpha=0 , heartbeats raise while typing answer.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Absolutely that's why I introduced π \pi there...... :-)

Rishabh Jain - 5 years, 2 months ago

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