It's an ∞ \red{\infty} sum!

Algebra Level 2

∑ n = 1 ∞ n n 4 + n 2 + 1 = ? \sum_{n=1}^{\infty}\dfrac{n}{n^4+n^2+1}=?


The answer is 0.5.

Zakir Husain by Zakir Husain , 41, India

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1 solution

Chew-Seong Cheong
Aug 27, 2020

S = ∑ n = 1 ∞ n n 4 + n 2 + 1 = ∑ n = 1 ∞ n ( n 2 − n + 1 ) ( n 2 + n + 1 ) = ∑ n = 1 ∞ 1 2 ( 1 n 2 − n + 1 − 1 n 2 + n + 1 ) = 1 2 ∑ n = 1 ∞ ( 1 n 2 − n + 1 − 1 ( n + 1 ) 2 − ( n + 1 ) + 1 ) = 1 2 ( ∑ n = 1 ∞ 1 n 2 − n + 1 − ∑ n = 2 ∞ 1 n 2 − n + 1 ) = 1 2 ( 1 1 2 − 1 + 1 ) = 1 2 = 0.5 \begin{aligned} S & = \sum_{n=1}^\infty \frac n{n^4+n^2+1} \\ & = \sum_{n=1}^\infty \frac n{(n^2-n+1)(n^2+n+1)} \\ & = \sum_{n=1}^\infty\frac 12 \left( \frac 1{n^2-n+1} - \frac 1{n^2+n+1} \right) \\ & = \frac 12 \sum_{n=1}^\infty \left( \frac 1{n^2-n+1} - \frac 1{(n+1)^2-(n+1)+1} \right) \\ & = \frac 12 \left(\sum_\blue{n=1}^\infty \frac 1{n^2-n+1} - \sum_\red{n=2}^\infty \frac 1{n^2-n+1} \right) \\ & = \frac 12 \left(\frac 1{1^2-1+1} \right) = \frac 12 = \boxed{0.5} \end{aligned}

3 Helpful 2 Interesting 4 Brilliant 0 Confused

@Chew-Seong Cheong I knew someone will definitely use telescopic series!

Zakir Husain - 9 months, 2 weeks ago

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I could not find another way.

Chew-Seong Cheong - 9 months, 2 weeks ago

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