Convergence Proof for Continuous Functions with Second Derivative at Zero

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Homework Statement



let f be a continuous function on an interval around 0, and let an=f(1/n) (for large enough n). prove that if f''(0) exists and f'(0)=f(0)=0, then
conv-1.jpg
converges

Homework Equations



i proved earlier in the problem that if the series converges, f(0)=0, and if f'(0) exists and the series converges, then f'(0)=0

The Attempt at a Solution


not really sure how to approach this. don't know if a convergence test should be used, but i know that lim n->infinity an=0 (zero test is inconclusive). what should i do?
 
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so you know
f(0) = 0

f'(0) = lim x->0, f(x)/x = 0

f''(0) = lim x->0, f'(x)/x = a

the 2nd line shows, f(x) tends to zero faster than x...
 
i'm not really sure how that helps. that takes me back to the result of the zero test, and the fact that f'(0)=0, which i already proved. how does that pertain to the series.
 
how about working towards a comparison test...? as f(x)/x tends to zero, near 0, maybe you can do something with f(1/n)/(1/n)
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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