Ask
your own question, for FREE!
Mathematics
10 Online
On finding the interval of convergence for a power series, I need help testing an endpoint. Endpoint x= -1 for ∑(1/n2^n)*(x-1)^n (which would be -2^n/n2^n) ....what test would I use to test the endpoint? Please show work.
Still Need Help?
Join the QuestionCove community and study together with friends!
\[\sum{ \frac{ (-2)^n }{ n2^n }} = \sum{ \frac{ (-1)^n2^n }{ n2^n }}=\sum{ \frac{ (-1)^n }{ n }}\] and then use the alternating series test from here, does that help?
IT would be 1/n? so it diverges???
Note the \[(-1)^n\] which makes it alternating.
so the lim as n goes to infinity of (-1)^n/n would be 0....so it converges?
yes, it converges but you must show that with the alternating series test. Do you know that test?
Still Need Help?
Join the QuestionCove community and study together with friends!
yes, it converges if it is a decreasing sequence and the lim equals zero
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
VERNEISH:
Two objects are moving along separate linear paths where each path is described by position, d, and time, t.
Maci:
what are the best stock apps for people who want to start investing?
Twaylor:
genuienly how do i get better self control i keep watcing youtube and theres this
IzzyCookie:
|dw:1792074056138:dw| Uhm, I don't know where to ask this, but I'm in college welding at my community college, and I need help learning vertical welding for
thehearken:
not sure if this is considered spam, im sorry if it is, but does anyone have tips
14 hours ago
1 Reply
0 Medals
5 days ago
3 Replies
2 Medals
1 week ago
3 Replies
0 Medals
3 days ago
9 Replies
1 Medal
1 week ago
2 Replies
1 Medal