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How to use the direct comparison test to evaluate?
The summation of e^-n^2 from 0 to infinity. The book says i must use the direct comparison method which i understand... Just not seeing which summation to compare this to, thanks :)
2 Answers
- kbLv 77 years agoFavorite Answer
This looks similar to a geometric series.
In this case, note that -n^2 ≤ -n for all n = 0, 1, 2, ... ,
Since Σ(n = 0 to ∞) e^(-n) is a convergent geometric series (with |r| = e^(-1) < 1), we conclude that the original series also converges.
I hope this helps!
- grunfeldLv 77 years ago
sigma( from 0 to inf ) [ e^( - n^2 ) ]
e^x = 1 + x + x^2 / 2! + x^3 / 3! + x^4 / 4! + x^5 / 5! + ............................
e^( - x^2 ) = 1 - x^2 + ( x^4 ) / 2! - (x^6) / 3! + ( x^8 ) / 4! - .....................
so sigma [ e^( - n^2 ) ]
= sigma( from 0 to inf ) [ 1 - x^2 + (x^4 ) / 2! - ( x^6 ) /3! + ( x^8 ) / 4! - ..................]
= DNE
this means that the series does not converge.
Source(s): my brain
