anonymous
  • anonymous
Find the values of x so that the series below converges. (((x+8)^(n))/(2^n)) please help find the interval. goes to INF and n=0
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
thomas5267
  • thomas5267
Is this the equation? \[ a_n=\frac{(x+8)^n}{2^n} \]
anonymous
  • anonymous
yes
thomas5267
  • thomas5267
What is the required limit?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I'm looking to find the values of x so that the series below converges. (the interval it does
anonymous
  • anonymous
limit is n=0
anonymous
  • anonymous
to INF
anonymous
  • anonymous
does that makes sense can u help?
thomas5267
  • thomas5267
So the question is find x such that: \[ \lim_{n\to0}\frac{(x+8)^n}{2^n}=\infty \]
anonymous
  • anonymous
no it has the simga logo in front of the fraction with a INF aboe th sigma and a n=0 below
anonymous
  • anonymous
\[\sum_{n=0}^{INF}\]
thomas5267
  • thomas5267
Find such x such that: \[ \sum_{n=0}^\infty\frac{(x+8)^n}{2^n}=\infty \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.