real analysis question

label Mathematics
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schedule 1 Day
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Oct 21st, 2017

If an > 0 and the series ∑n=0an diverges, then the partial sums sn = ∑k=0n ak →∞ as n →∞. Note also that sn > sn – 1 > 0 or all n.

Since an = sn – sn – 1,we conclude that an / sn2 = (sn – sn – 1) / sn2 < (sn – sn – 1) / sn sn – 1 =1/sn – 1 – 1/sn .

The series ∑n=1(1/sn – 1 – 1/sn) converges because its partial sums ∑k=1n (1/s k –1 – 1/sk) = 1/s 0 – 1/sn are bounded from above by 1/s 0 = 1/a 0. Hence, by the comparison principle the series ∑n=0an / sn2 also converges. 


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