How do you use the limit comparison test for #sum ((sqrt(n+1)) / (n^2 + 1))# as n goes to infinity?

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Answer 1 · 2015-04-14T18:40:54

Remembering that:

#sum_(n=0)^(+oo)1/n^alpha# is convergent if #alpha>1#, than:

#(sqrt(n+1)) / (n^2 + 1)~sqrt(n)/n^2=n^(1/2)/n^2=1/n^(3/2)#,

so that series is convergent.

Answer 2 · 2015-04-14T18:56:41

This series behaves like #sqrtn/n^2 = 1/n^(3/2)#

Compare it to that using the limit comparison test:

#(sqrt(n+1)/(n^2+1))/(1/(n^(3/2))) = (sqrt(n+1)/(n^2+1)) n^(3/2)/1=(n^(3/2)sqrt(n+1))/(n^2+1)#

#= (n^(3/2)sqrtn sqrt(1+1/n))/(n^2(1+1/n^2)) = (sqrt(1+1/n))/(1+1/n^2)#

The limit as #n rarr oo# is 1, which is positive.

Since #sum 1/n^(3/2)# converges (p-series), so does #sum(sqrt(n+1)/(n^2+1)#