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Let a_n be a non identically zero arithmetic progression. Then, is it true that Sum (1/a_n) diverges?

2007-09-04 06:37:58 · 2 answers · asked by Sonia 1 in Science & Mathematics Mathematics

2 answers

a_n = a_1 + r * (n-1)

= (n-1) * (r+a_1/(n-1))

If n is large, | r+a_1/(n-1) | < 2r.

So a_n < (n-1) * 2r for n large.

So 1/a_n > 1/(2r) * (1/(n-1))

So, there is a C = 1/(2r) such that 1/a_n > C/(n-1) for large n.

Since sum (1/(n-1)) diverges, sum(1/a_n) diverges.

2007-09-04 06:53:10 · answer #1 · answered by thomasoa 5 · 2 0

an= ao+n*r ( positive terms)
bn= 1/an is of the same class as 1/n which diverges

2007-09-04 07:41:48 · answer #2 · answered by santmann2002 7 · 0 0

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