An is positive
ΣAn converges
Ok but Raffaele is right.We cannot determine convergence or divergence
without having the An term because:
Σ(n = 0 to ∞) (-1)^n[1/n]
this is a convergent series as lim 1/n =0 (as n->∞)
and an+1 ≤ an
but if you take:
Σ(n = 0 to ∞) (-1)^n[n/(7n-6)]
then it is a divergent series as lim n/(7n-6) ≠ 0 (as n->∞)
something is missing
alternate series converge only if
lim An = 0 (n --> infinity)
sometimes called Leibnitz criterium
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Verified answer
Ok but Raffaele is right.We cannot determine convergence or divergence
without having the An term because:
Σ(n = 0 to ∞) (-1)^n[1/n]
this is a convergent series as lim 1/n =0 (as n->∞)
and an+1 ≤ an
but if you take:
Σ(n = 0 to ∞) (-1)^n[n/(7n-6)]
then it is a divergent series as lim n/(7n-6) ≠ 0 (as n->∞)
something is missing
alternate series converge only if
lim An = 0 (n --> infinity)
sometimes called Leibnitz criterium