How do you test for convergence of #Sigma (3n-7)/(10n+9)# from #n=[0,oo)#?

1 Answer
May 18, 2017

By putting an limit in front.

Explanation:

#lim_(n=>oo)((3n-7)/(10n+9))#
In this case, if the result is #!=0# there's divergence.
But the result is something that we cannot solve.
#=>[oo/oo]#
So we are going to isolate n in the equation:
#=>lim_(n=>oo)((n(3-7/n))/(n(10+9/n)))#
#=lim_(n=>oo)((3-7/n)/(10+9/n))#
And we just have to solve it!
#=(3-7/oo)/(10+9/oo)#
#=(3-0)/(10+0)#
#=3/10#
The result is #!=0# so there's divergence!
(I'not an english speaker so I don't know the name of this method. Feel free to add it.)