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

Question

Evelyn Agard · Accepted Answer

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.

Elijah Abram · Answer

#sum_(n=1)^oo1/(n+sqrt(n))# diverges, this can be seen by comparing it to #sum_(n=1)^oo1/(2n)#.

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.

Isabella Ackerman · Answer

undefined To use the limit comparison test for the series $\sum_{n=1}^{\infty} \frac{1}{n + \sqrt{n}}$, we need to compare it with a known series whose convergence behavior is established. We can choose to compare it with the harmonic series, $\sum_{n=1}^{\infty} \frac{1}{n}$, since it is a commonly known series. By taking the limit as $n$ approaches infinity of the ratio of the terms of the given series and the terms of the harmonic series, we can determine if the given series converges or diverges. If the limit is a positive finite number, then both series behave similarly, and if it's zero or infinity, then the behavior of the series differs.