Ratio test to solve infinte Series converges

Evaluate ` sum_(n=1)^(oo)(1-3/n)^n ` :

We are asked to use the ratio test to test for convergence. If the limit of the ratio of the (n+1)st term to the nth term is less than 1 in absolute value, the series converges. If greater than 1, the series diverges. (If the limit equals 1, we cannot draw any conclusion.)

`lim_(n->oo)(1-3/(n+1))^(n+1)/(1-3/n)^n `

`=lim_(n->oo)((n+1)ln(1-3/(n+1)))/(nln(1-3/n)) ` Using L'Hopital's rule for the indeterminant form

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

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

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

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

`=lim_(n->oo)(1-3/n)/(1-1/n-2/n^2)=1 `

Thus the ratio test is indeterminate.

(Using the limit test we can show that the series diverges.)

Comments

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