How do you use the integral test to determine if #Sigma e^(-n/2)# from #[1,oo)# is convergent or divergent?

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Answer 1 · 2017-10-26T12:54:57

The sum #sum_1 ^ooe^(-2/n)# is convergent

We must calculate the improper integral

#int_1^ooe^(-x/2)dx#

First we compute the integral

#I=inte^(-x/2)dx#

Perform the substitution

#u=-x/2#, #=>#, #du=-(dx)/2#

Therefore,

#I=int-2e^udu=-2e^u=-2e^(-x/2)+C#

Now, compute the boundaries

#lim_(x->1^+)-2e^(-x/2)=-2e^(-1/2)=-2/sqrte#

#lim_(x->+oo)-2e^(-x/2)=0#

Therefore,

#int_1^ooe^(-x/2)dx=2/sqrte#

The integral is convergent.