Recent content by atthebeach
-
A
Convergence of the Sequence An = √(2^n + 3^n)
OHHH the limit would be three thank you so much. I think maybe i was supposed to do an epsilon proof but this should be sufficient- atthebeach
- Post #13
- Forum: Calculus and Beyond Homework Help
-
A
Convergence of the Sequence An = √(2^n + 3^n)
yes. but then i have the 1 under the radical and i know that the infinith root of 1 would be 1. but...i cannot split up the terms under the radical- atthebeach
- Post #11
- Forum: Calculus and Beyond Homework Help
-
A
Convergence of the Sequence An = √(2^n + 3^n)
oops i meant to write that. so from here do i need to find the N?- atthebeach
- Post #9
- Forum: Calculus and Beyond Homework Help
-
A
Convergence of the Sequence An = √(2^n + 3^n)
I think I figured it out. would it be 3 * nth root of ((3/2)^n+1)?- atthebeach
- Post #7
- Forum: Calculus and Beyond Homework Help
-
A
Convergence of the Sequence An = √(2^n + 3^n)
It is monotone decreasing. Looking at a graph I can deduce that it converges to 3. but I really struggle with the proof...and the algebra sadly. How would I factor 3^n out of the radical??- atthebeach
- Post #4
- Forum: Calculus and Beyond Homework Help
-
A
Convergence of the Sequence An = √(2^n + 3^n)
If an = \sqrt[n]{2^n+3^n} does the sequence converge? Prove your assertion. I have no idea where to start with this problem. It does have something to with \exists N such that n>N \Rightarrow |an - a| < ϵ for all ϵ>0 Can yall help me? It would be greatly appreciated.- atthebeach
- Thread
- Sequence
- Replies: 13
- Forum: Calculus and Beyond Homework Help