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January 30, 2015

January 30, 2015

Posted by **sarah** on Thursday, February 21, 2008 at 6:41pm.

1. if the lim as n->infinity of a(sub n)=0, then the sum from n=1 to infinity of a(sub n) converges

i said this was true because I know that if a (sub n) does NOT=0, it diverges

2. if the sum from n=1 to infinity of a(sub n) converges and a(sub n) does not =0, then the sun from n=1 to infinity of 1/(a(sub n)) diverges.

?????

- calculus -
**Count Iblis**, Thursday, February 21, 2008 at 6:53pm1) is false. Counterexample: a_n = 1/n

Try to prove that sum from n=1 to infinity of 1/n is divergent.

"i said this was true because I know that if a (sub n) does NOT=0, it diverges"

Which is logically equivalent to:

Not divergent implies a_n ---> 0.

Of course, a convergent series must be such that a_n --->0. But the reverse is not true. So, the condition a_n ---> 0 is necessary but not sufficient for convergence.

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