Asked by ANON
Does the series converge or diverge? Use the comparison test. Thank you
((1/n)-(1/n^2))^n
((1/n)-(1/n^2))^n
Answers
Answered by
Steve
a simple convergent geometric series is r^n where r<1.
(1/n - 1/n^2) = (n-1)/n^2 < 1 for all n>0
In fact,
1/n (1 - 1/n) < 1/n < 1/2 for n>2.
So, An < 1/2^n for n>2
(1/n - 1/n^2) = (n-1)/n^2 < 1 for all n>0
In fact,
1/n (1 - 1/n) < 1/n < 1/2 for n>2.
So, An < 1/2^n for n>2
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