Find the formula to express the infinite series...

  • #1
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The problem is to find the general term ##a_n## (not the partial sum) of the infinite series with a starting point n=1

$$a_n = \frac {8} {1^2 + 1} + \frac {1} {2^2 + 1} + \frac {8} {3^2 + 1} + \frac {1} {4^2 + 1} + \text {...}$$

The denominator is easy, just ##n^2 + 1## but I can't think of any way to get the numerator to alternate between 8 and 1.
 

Answers and Replies

  • #2
22,089
3,296
##4.5 + 3.5(-1)^{n+1}##
 
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  • #3
21
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Never mind I figured it out $$a_n = \frac {9 - 7 \{ -1 \} ^n} {2 \{ 1 + n^2 \}}$$
 
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