Reciprocals???

Algebra Level 2

If ∣ a ∣ > 1 | a | >1

1 a + 1 a 2 + 1 a 3 + . . . . ∞ = ? \dfrac{1}{a}+\dfrac{1}{a^2}+\dfrac{1}{a^3}+....\infty = ?

a Depends on a 1/a-1 1 1/a

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Nicholas Patrick by Nicholas Patrick , 21, Indonesia

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1 solution

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For this to be true, ∣ a ∣ > 1 |a|> 1 which is not given in question.

Still, 1 a − 1 \dfrac{1}{a-1} depends on a a . Isn't it?

Pranjal Jain - 6 years, 4 months ago

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divergen??

Aswad Hariri Mangalaeng - 6 years, 4 months ago

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Yeah! Series diverges for ∣ a ∣ ≤ 1 |a|\leq 1

Pranjal Jain - 6 years, 4 months ago

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