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