How do you use the limit comparison test for #sum 1 / (n + sqrt(n))# for n=1 to #n=oo#?

Since this series is a sum of positive numbers, we need to find either a convergent series #sum_(n=1)^(oo)a_n# such that #a_n>=1/(n+sqrt(n))# and conclude that our series is convergent, or we need to find a divergent series such that #a_n<=1/(n+sqrt(n))# and conclude our series to be divergent as well.

We remark the following:
For
#n>=1#, #sqrt(n)<=n#.
Therefore
#n+sqrt(n)<=2n#.
So
#1/(n+sqrt(n))>=1/(2n)#.

Since it is well known that #sum_(n=1)^oo1/n# diverges, so #sum_(n=1)^oo1/(2n)# diverges as well, since if it would converge, then #2sum_(n=1)^oo1/(2n)=sum_(n=1)^oo1/n# would converge as well, and this is not the case.

Now using the comparison test, we see that #sum_(n=1)^oo1/(n+sqrt(n))# diverges.

The limit comparison test takes two series, #suma_n# and #sumb_n# where #a_n>=0#, #b_ngt0#.

If #lim_(nrarroo)a_n/b_n=L# where #L>0# and is finite, then either both series converge or both series diverge.

We should let #a_n=1/(n+sqrtn)#, the sequence from the given series. A good #b_n# choice is the overpowering function that #a_n# approaches as #n# becomes large. So, let #b_n=1/n#.

Note that #sumb_n# diverges (it's the harmonic series).

So, we see that #lim_(nrarroo)a_n/b_n=lim_(nrarroo)(1/(n+sqrtn))/(1/n)=lim_(nrarroo)n/(n+sqrtn)#. Continuing by dividing through by #n/n#, this becomes #lim_(nrarroo)1/(1+1/sqrtn)=1/1=1#.

Since the limit is #1#, which is #>0# and defined, we see that #suma_n# and #sumb_n# will both diverge or converge. Since we already know at #sumb_n# diverges, we can conclude that #suma_n=sum_(n=1)^oo1/(n+sqrtn)# diverges as well.