how to get this result
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I would try to get the the formula for the finite sum (say n = 0 ... m) and then take the limit as m ---> infinity. I have not tried it yet, but I would factor the denominator (use the quadratic formula) into 2 linear factors and then break it up into 2 sums using partial fractions. This usually turn out to be a "telescoping" series. I'll let you know how I made out. Oh, and BTW, when n = 0 then S1 = 1 so would start with n = 1 ... m and then just add the 1 at the end.
You can see the sqrt(5) coming as b^2-4ac = 9-4 = 5 so that's encouraging. I'll let you know how I make out.
Been at this one for a while - no breakthrough yet. Maybe completing the square. The sqrt(5) in the answer sure looks like it is related somehow. We know this series has to converge since the sum(n = 1--> infinity) of [1/n^2] = (pi)^2/6 and this series is smaller term by term.
I factored it into linear factors and tried partial fractions and you get an alternating series but no luck.
@Ian Fowler ok i found an answer
@ian fowler, can i see your alternating series, i might be able to squeeze an answer out of that