How do you find the interval of convergence $Sigma ((-1)^n(x+2)^n)/n$ from $n=[1,oo)$ ?

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How do you find the interval of convergence $Sigma ((-1)^n(x+2)^n)/n$ from $n=[1,oo)$ ?

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Answer 1 · 2017-01-24T22:57:56

The series:

$sum_(n=1)^oo (-1)^n (x+2)^n/n$

is absolutely convergent for $x in (-3,-1)$ and simply convergent for $x =-1$

Explanation:

To find the interval of convergence we can apply the ratio test, stating that a necessary condition for a series $sum_(n=1)^oo a_n$ to converge is that:

$L = lim_(n->oo) abs (a_(n+1)/a_n) <= 1$

If $L < 1$ the condition is also sufficient and the series converges absolutely.

Let's calculate the ratio:

$abs (a_(n+1)/a_n) = abs (frac ( (x+2)^(n+1)/(n+1)) ((x+2)^n/n)) = abs(x+2) n/(n+1)$

so that:

$lim_(n->oo) abs (a_(n+1)/a_n) = abs(x+2)$

we can therefore conclude that the series is absolutely convergent for $abs (x+2) < 1$ and divergent for $abs (x+2) > 1$ .

In the case where $abs (x+2) = 1$ we know that the test is inconclusive but we can see that for $x=-3$ the series becomes:

$sum_(n=1)^oo (-1)^n(-1)^n/n = sum_(n=1)^oo 1/n$

that is divergent, while for $x= -1$ we have:

$sum_(n=1)^oo (-1)^n/n$

that is convergent, but not absolutely convergent.

In conclusion the series converges for $x in (-3,-1]$

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How do you find the interval of convergence $Sigma ((-1)^n(x+2)^n)/n$ from $n=[1,oo)$ ?

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Explanation:

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How do you find the interval of convergence Sigma ((-1)^n(x+2)^n)/nSigma ((-1)^n(x+2)^n)/n from n=[1,oo)n=[1,oo)?

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How do you find the interval of convergence $Sigma ((-1)^n(x+2)^n)/n$ from $n=[1,oo)$ ?