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By using the Ratio Test determine if the series converges: `Sigma` `(-1)^n/4^n`

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Rocky52 | Student, Undergraduate | Honors

Posted April 28, 2013 at 4:53 AM via web

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By using the Ratio Test determine if the series converges:

`Sigma` `(-1)^n/4^n`

Tagged with calculus 2, math, series

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Rocky52 | Student , Undergraduate | Honors

Posted April 28, 2013 at 5:05 AM (Answer #1)

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By using The Ratio Test determine if the series converges:

`sum` `(-1)^n/4^n`

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rakesh05 | High School Teacher | (Level 1) Assistant Educator

Posted April 28, 2013 at 5:31 AM (Answer #2)

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 The series `sum_(n=1)^ooa_n`  where `a_n`  is non zero is

(i) convergent if the ratio  `lim_(n->oo)mod(a_(n+1)/a_n)<1`

(ii)divergent if the ratio `lim_(n->oo)mod(a_(n+1)/a_n)>1` .

Here `a_n=(-1)^n/4^n` ,      `a_(n+1)=(-1)^(n+1)/4^(n+1)`

Now  `a_(n+1)/a_n={(-1)^(n+1)/4^(n+1)}/{(-1)^n/4^n}` =-1/4

Now   `lim_(n->oo)mod (a_(n+1)/a_n)=1/4<1`

So the given series is convergent.

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tiburtius | High School Teacher | (Level 3) Associate Educator

Posted April 28, 2013 at 5:38 AM (Answer #3)

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This is an alternating series. If alternating series is absolutely convergent (series of absolute values is convergent) then the alternating series is convergent` `

`sum |a_n|->a=>sum a_n->b,` `a,b in RR`

 

Also if `lim_(n->oo)|(a_(n+1))/(a_n)|=q` then the series is convergent if `q<1` and divergent if `q>1.`

Let's now check convergence of our series.

`lim_(n->oo)|((-1)^(n+1))/(4^(n+1))/((-1)^n)/(4^n)|=lim_(n->oo)4^n/4^(n+1)=`

`lim_(n->oo)4^n/(4cdot4^n)=lim_(n->oo)1/4=1/4`

Since `1/4<1` it follows that the series is convergent.

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