Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

I can already feel my brain shutting down ;)
mine shutdown like 3 hours ago
1 Attachment
Part A's easy

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

yes...i get 65/24
1+1+1/2+1/6+1/24
YAY :D
part b looks like induction
part b is definitely induction ... this property could be used to show that 2
for n=1 we have 1/1!=1 and 1/2^(1-1)=1/1=1 \(1\le 1\) so the basis step holds assume \[\frac{1}{k!}\le\frac{1}{2^{k-1}}\] is true for some \(k\) then \[\frac{1}{(k+1)!}=\frac{1}{k!}\frac{1}{k+1}\le\frac{1}{2^{k-1}}\frac{1}{k+1}\] since \(\frac{1}{2}\ge\frac{1}{n+1}\) for \(n\ge1\) we have \[\frac{1}{2^{k-1}}\frac{1}{k+1}\le\frac{1}{2^{k-1}}\frac{1}{2}=\frac{1}{2^{k}}\] QED
Prof Zarkon I need your inputs on another problem, i have pinged you from the relevant thread. Please do check it :)
btw for part c the upper bound is \( 3\).
ty guys, i can die a peaceful death now, knowing that 9.3 is finally done...
when (exactly) are you going to die? :P
not any time soon, hopefully.

Not the answer you are looking for?

Search for more explanations.

Ask your own question