# calculus II

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This image has been Flagged as inappropriate Click to unflag 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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