+0  
 
0
129
2
avatar

(sqrt(2)/e^2)+(sqrt(2)/e^4)+(sqrt(2)/e^6)+(sqrt(2)/e^8)+...

 

Not sure if I started it out correctly, but I turned it into a geometric series:

infinity ∑ n=1 (sqrt(2)/e^2)^n

 

a=1 & r = (sqrt(2)/e^2)

 

& if I'm right if |r| < 1, then it's convergent. If |r| ≥ 1.

Guest May 3, 2017
Sort: 

2+0 Answers

 #1
avatar
0

The sqrt(2) doesn't increase in power term by term, (i.e it stays at sqrt(2) for each term).

Remove it as a common fact and what is left is a GP.

Guest May 3, 2017
 #2
avatar
0

The series should converge to: sqrt(2)/((e - 1) (1 + e)) = ~0.221349373.....etc.

Guest May 3, 2017

8 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details