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JG89
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Homework Statement
Prove that if the infinite series a_1 + a_2 + ... + a_n, with positive decreasing terms, converges to a value A, then the limit as n approaches infinity of n*a_n = 0
Homework Equations
The Attempt at a Solution
Assume [tex] \forall n \in Z^{+} ,\exists \epsilon > 0: n(a_n) \ge \epsilon [/tex], where Z+ is the set of all positive integers and of course, epsilon is fixed. Now, if n(a_n) >= epsilon, then a_n >= epsilon/n
If epsilon >= 1, then a_n >= epsilon/n >= 1/n, for all n. But this can't be true because then
a_1 >= 1/1, a_2 >= 1/2, a_3 >= 1/3 and so on, implying that a_1 + a_2 + ... + a_n >= 1/1 + 1/2 + 1/3 + 1/4 + ... = harmonic series, which diverges, and so a_n diverges, but a_n really converges.
Now we consider 0 < epsilon < 1:
There exists a fixed positive number c such that c*epsilon = 1. If n(a_n) >= epsilon, then
cn(a_n) >= c*epsilon = 1, implying that c(a_n) >= 1/n for all positive integers n.
And so, c(a_1) >= 1/1, c(a_2) >= 1/2, c(a_3) >= 1/3 and so on, implying:
c(a_1) + c(a_2) + c(a_3) = c(a_1 + a_2 + ..) >= 1/1 + 1/2 + 1/3 + ... = harmonic series, which diverges, and so the infinite series c(a_1) + c(a_2) + .. must also diverge. However, c(a_1 + a_2 + ...) converges to the value cA, since c is a fixed value.
These two contradictions both lead to the conclusion that there doesn't exist an epsilon such that n(a_n) >= epsilon for all positive integers n, meaning that n(a_n) < epsilon for all positive epsilon and for n large enough, meaning that the limit is 0.
QED
This proof looks fine to me, but the solution seems too simple. What is wrong with the proof?
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