sum n=1 to Infinity (9)^n-1/2^n, converges or diverges
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
The numerator far outpaces the denominator as n increases.
what kind of method or test should I use for this question?
Not the answer you are looking for? Search for more explanations.
For most fractional problems like this you can do what's called the ratio test where you take the limit of the series function at infinity, and if it's greater than zero it diverges, less than zero it converges and equal to zero it is inconclusive. L'Hopital's rule can be used in most situations where you end up with ∞/∞ or 0/0.
So for this one, take the limit at infinity and you end up with 9^(∞-1)/2^∞.
That's essentially 9^∞/2^∞ since the -1 isn't really making a dent in infinity.
you can safely take the ∞th-root of the entire expression without changing its sign, so that's
9/2 and since 9/2 > 1, it's divergent by the ratio test.
Slight correction to my previous note: We're not worried about changing the limit's sign, but in making it cross between being greater or less than 1. Taking a root is still safe.