If
a_{n}
> 0 and the series ∑_{n=0}^{∞
}a_{n}
diverges, then the partial sums s_{n}
= ∑_{k=0}^{n
}a_{k}
→∞
as n →∞.
Note also that s_{n}
> s_{n – 1}
> 0 or all n.

The
series ∑_{n=1}^{∞
}(1/s_{n
– 1 }– 1/s_{n})
converges because its partial sums ∑_{k=1}^{n
}(1/s_{
k –1 }– 1/s_{k})
= 1/s_{
0 }– 1/s_{n
}are bounded from
above by 1/s_{
0 }= 1/a_{
0}. Hence, by the
comparison principle the series ∑_{n=0}^{∞
}a_{n
}/ s_{n}^{2
}also converges.

Apr 9th, 2015

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