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Is the series #\sum_(n=1)^\infty((-5)^(2n))/(n^2 9^n)# absolutely convergent, conditionally convergent or divergent?

(Use the appropriate test)

Answer 1

Emilia Aguilar

The given series (\sum_{n=1}^\infty \frac{(-5)^{2n}}{n^2 9^n}) is absolutely convergent.

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Answer 2

Scarlett Allison

Diverges by the Ratio Test.

We'll check for Absolute Convergence using the Ratio Test, where #a_n=(-5)^(2n)/(n^2 9^n)#.

The Ratio Test tells us we take

#L=lim_(n->oo)|a_(n+1)/a_n|,# and

#L<1# implies absolute convergence (and hence convergence),

#L>1# implies divergence,

#L=1# is inconclusive and we need to use another test.

Then,

#a_(n+1)=(-5)^(2(n+1))/((n+1)^2 9^(n+1))=(-5)^(2n+2)/((n+1)^2 9^(n+1))#

Thus,

#L=lim_(n->oo)|(-5)^(2n+2)/((n+1)^2 9^(n+1))*(n^2 9^n)/(-5)^(2n)|#

#(-5)^(2n+2)/(-5)^(2n)=(-5)^2=25#

#9^n/9^(n+1)=1/9#

We can factor these constants outside of the limit, getting

#25/9lim_(n->oo)n^2/(n+1)^2=25/9>1#

We dropped the absolute value bars as everything is positive when we go to infinity.

We also then have divergence by the Ratio Test.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.