Sum of a Series - Part 2

Calculus Level 5

Let { a n } \{ a_n \} be a sequence defined as a 0 = a 1 = 1 a_0 = a_1 = 1 and a n + 2 = 2 a n + 1 + a n − n a_{n+2} = 2 a_{n+1} + a_n - n for n ≥ 0 n \ge 0 .

Find the value of this infinite summation S = ∑ k = 0 ∞ a k 3 k S = \sum_{k=0}^{\infty} \frac{a_k}{3^k}


The answer is 2.625.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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1 solution

Otto Bretscher
Apr 30, 2016

We can adapt our solution here .

Consider the generating function A ( x ) = ∑ a n x n = 1 + x + ∑ a n + 2 x n + 2 A(x)=\sum a_nx^n=1+x+\sum a_{n+2}x^{n+2} = 1 + x + 2 ∑ a n + 1 x n + 2 + ∑ a n x n + 2 − ∑ n x n + 2 = 1 + x + 2 x ( A ( x ) − 1 ) + x 2 A ( x ) − x 3 ( x − 1 ) 2 =1+x+2\sum a_{n+1}x^{n+2}+\sum a_nx^{n+2}-\sum nx^{n+2}=1+x+2x(A(x)-1)+x^2A(x)-\frac{x^3}{(x-1)^2} . We solve and find A ( x ) = ( 2 x − 1 ) ( x 2 − x + 1 ) ( x − 1 ) 2 ( x 2 + 2 x − 1 ) A(x)=\frac{(2x-1)(x^2-x+1)}{(x-1)^2(x^2+2x-1)} for ∣ x ∣ < 2 − 1 |x|<\sqrt{2}-1 . For x = 1 3 x=\frac{1}{3} we have S = A ( 1 3 ) = 2.625 S=A\left(\frac{1}{3}\right)=\boxed{2.625}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Isn't it kinda surprising that the difference between the two summations is just ≈ 0.4 \approx 0.4 ?

A Former Brilliant Member - 5 years, 1 month ago

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