calculus II

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a) You need to use the ratio test to evaluate the radius of the convergence of the given sequence, hence, you need to evaluate the limit `lim_(n->oo) |(a_(n+1))/(a_n)|` , where `a_n = (x^n*(n+1))/(3^n)` such that:

`lim_(n->oo) |((n+2)x^(n+1))/(3^(n+1))*(3^n/((n+1)x^n))|`

Reducing duplicate terms yields:

`lim_(n->oo) |((n+2)x)/3)*(1/(n+1))|`

You need to factor out `x/3,` such that:

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a) You need to use the ratio test to evaluate the radius of the convergence of the given sequence, hence, you need to evaluate the limit `lim_(n->oo) |(a_(n+1))/(a_n)|` , where `a_n = (x^n*(n+1))/(3^n)` such that:

`lim_(n->oo) |((n+2)x^(n+1))/(3^(n+1))*(3^n/((n+1)x^n))|`

Reducing duplicate terms yields:

`lim_(n->oo) |((n+2)x)/3)*(1/(n+1))|`

You need to factor out `x/3,` such that:

`|x|*lim_(n->oo) ((n+2))/3)*(1/(n+1)) = |x/3|*lim_(n->oo) (n(1 + 2/n))/(n(1 + 1/n))`

Reducing duplicate factors yields:

`|x/3|*lim_(n->oo) (1 + 2/n)/(1 + 1/n) = |x/3|*1 = |x/3|`

By the ratio test yields:

- if `|x/3| < 1 => |x| < 3` , the series converges

- if `|x| > 3` , the series diverges

Hence the radius of convergence of series is `R = 3.`

b) You need to evaluate the interval of convergence, hence, you need to solve the absolute value inequality `|x| < 3` , such that:

`|x| < 3 => -3 < x < 3`

Hence, evaluating the interval of convergence yields `(-3,3).`

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