Calculus II

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The series consists of the terms `-2/3, 4/9, -6/27, 8/81, -10/243...`

The nth term of the series is `((-1)^n*2*n)/3^n`

It has to be determined if `sum_(n=1)^oo ((-1)^n*2*n)/3^n` converges, diverges or does neither of the two.


= `lim_(n->oo)(1^n*2*n)/3^n`

`1^oo` , is indeterminate, this allows the use of l'Hospital's rule and the numerator and denominator can be replaced with their derivatives.

`lim_(n->oo) 2/(3^n*ln 3)`

= 0

As `lim_(n->oo) (-1^n*2*n)/3^n` tends to a constant number, the series `sum_(n=1)^oo (-1^n*2*n)/3^n` is also convergent

The series `sum_(n=1)^oo ((-1)^n*2*n)/3^n` is convergent.

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