blackstreet23
  • blackstreet23
I am trying to do an alternating series test problem but i forgot how to prove if the series is decreasing. I remember they were several tests for that, but i forgot which where. Please help!
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
freckles
  • freckles
Can you show me your series?
freckles
  • freckles
Sometimes it is pretty easy with derivative. Sometimes it is easy with algebra. like we can show \[a_{n+1}
freckles
  • freckles
n not x

Looking for something else?

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

More answers

blackstreet23
  • blackstreet23
could you tell me all the tests? I am trying to review my calc 2 to be ready for calc 3
freckles
  • freckles
Those are the only two I can recall.
SolomonZelman
  • SolomonZelman
In alternating series, if the absolute value of the terms decreases then the series will converge.
blackstreet23
  • blackstreet23
1 Attachment
SolomonZelman
  • SolomonZelman
So, I will denote the sums with \(\rm S_1,~S_2...\) |dw:1441594022749:dw|
blackstreet23
  • blackstreet23
that is what i have so far
SolomonZelman
  • SolomonZelman
Can you post it in here, more clearly.....
freckles
  • freckles
Is the attachment bringing up a new question?
blackstreet23
  • blackstreet23
no just showing my work so far
freckles
  • freckles
\[k+1>k \text{ so } \frac{1}{k+1}<\frac{1}{k}\]
freckles
  • freckles
or. ... \[f(k)=\frac{1}{k+1} \\ f'(k)=\frac{-1}{(k+1)^2} \\ \frac{-1}{(k+1)^2}<0 \text{ for all } k \text{ except } k=-1 \\ \text{ which we aren't concerned about }\] either one of these ways shows a_k is a decreasing sequence
SolomonZelman
  • SolomonZelman
\(\large\color{black}{ \displaystyle \sum_{ k=0 }^{ \infty } \frac{x^k}{k+1}}\) \(\large\color{black}{ \displaystyle \lim_{ k\rightarrow \infty }\frac{x^{k+1}(k+1)}{(k+2)x^k}}\) \(\large\color{black}{ \displaystyle x\lim_{ k\rightarrow \infty }\frac{k+1}{k+2}}\) \(\large\color{black}{ \displaystyle|x |<1 }\)
SolomonZelman
  • SolomonZelman
and it is actually good, because even if it was just x\(^k\) without dividing by k+1, then it converges for |x|<1. And when you divide by k+1, then it is even smaller. (And x=-1 is included in convergence, because it is alternating harmonic series when x=-1)
SolomonZelman
  • SolomonZelman
So it should really be: \(-1 \le x<1\)

Looking for something else?

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