Recent content by Raymondyhq

I could set up the equation (11/10)n = n+1 but I would not be able to solve it without the aid of a graphing calculator. Using a graphing calculator, I found that for all x > 40, pn is greater than n+1. Thus, I would have proved that when n goes from 1 to positive infinity, the series (33/40)n >...

What about p = 11/10 ? pn would be greater than (n+1) if n is sufficiently large, and pn * (3/4)n = (33/40)n which would satisfy r < 1.

In this case the quotient would need to be ... greater than 1?

I am having trouble finding a p that satisfies pn>(n+1) :cry:

This is exactly what I am having trouble solving. I could try to move n into the brackets with 3/4. ∑n(3/4)n = ∑(n√n)n(3/4)n = ∑(n√n 3/4)n Then I am stuck.

Could you elaborate? I don't quite understand what you mean by ∑qn

Homework Statement Prove the convergence of this series using the Comparison Test/Limiting Comparison Test with the geometric series or p-series. The series is: The sum of [(n+1)(3^n) / (2^(2n))] from n=1 to positive ∞ The question is also attached as a .png file 2. Homework Equations The...

Let us assume that d is a vector in the vector space ℝ2 , then is: {td | t ∈ ℝ} the same as span{d} ? Thank you.