An infinite series

Algebra Level 3

1 1 + 1 2 + 1 4 + 2 1 + 2 2 + 2 4 + 3 1 + 3 2 + 3 4 + \dfrac{1}{1+1^2 + 1^4} +\dfrac{2}{1+2^2 +2^4}+\dfrac{3}{1+3^2 +3^4}+ \ldots


The answer is 0.5.

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1 solution

Sandeep Rathod
Nov 16, 2014

T n = n 1 + n 2 + n 4 T_{n} = \frac{n}{1 + n^{2} + n^{4}}

T n = n 1 + 2 n 2 + n 4 n 2 T_{n} = \frac{n}{1 + 2n^{2} + n^{4} - n^{2}}

T n = n ( 1 + n 2 ) 2 n 2 T_{n} = \frac{n}{(1 + n^{2})^{2} - n^{2}}

= 1 2 ( 1 n 2 n + 1 1 n 2 + n + 1 ) = \frac{1}{2}(\frac{1}{n^{2} - n + 1} - \frac{1}{n^{2} + n + 1})

Since n n tends to \infty the answer becomes 1 2 \boxed{\frac{1}{2}}

Edited for clarity.

Krishna Ar - 6 years, 6 months ago

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OK Thank you

sandeep Rathod - 6 years, 6 months ago

i did the same way!!

Aareyan Manzoor - 6 years, 5 months ago

I did in the same way

Sudhir Aripirala - 5 years, 11 months ago

nice solution

Faheem Mubarak - 1 year, 2 months ago

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