- #1

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What I tried was finding how many terms it would take the have an error that was < .0001 then found the sum with that many terms... I got 0.10969 as the partial sum using 4 terms.

- Thread starter Nick_L
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- #1

- 7

- 0

What I tried was finding how many terms it would take the have an error that was < .0001 then found the sum with that many terms... I got 0.10969 as the partial sum using 4 terms.

- #2

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[tex]\sum_{n=1}^{\infty}nx^n=\frac{x}{(x-1)^2}[/tex]

which can be obtained from the geometric series by computing the derivative and multiplying by x. Hence

[tex]\sum_{n=1}^{\infty}n(-\frac{1}{11})^n=-\frac{11}{144}[/tex]

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