Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

See more answers at brainly.com
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
I have never gotten to series in calculus Lala, sorry.. but I called someone better for you :)
In a geometric series, the series will converge if \[\large |r|<1\]In this case, \[|-\frac{1}{2}| <1 \implies \frac{1}{2}< 1\] therefore the \[\large \sum_{n-1}^{\infty}-4(-\frac{1}{2})^{n-1} = C\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

thank you @Luigi0210
Because it converges, your sum : \[\large S=\frac{a}{1-r}\] where \[\large r=-\frac{1}{2}\]
that stuff looks so familiar >.<
TRue that
so that is the answer?
yes
oh ok!
thanks
you think you could help with another?

Not the answer you are looking for?

Search for more explanations.

Ask your own question