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

Question

Isaiah Allen · Accepted Answer

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.

Emily Ammerman · Answer

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)#

Evelyn Agard · Answer

undefined To use the limit comparison test for the series $ \sum \frac{\sqrt{n+1}}{n^2 + 1} $ as $ n $ goes to infinity, you need to find a series $ \sum b_n $ that you can compare it to. Here's how you do it:

1. Identify the problematic part of the series. In this case, it's $ \sqrt{n+1} $ in the numerator.

2. Find a simpler series $ \sum b_n $ where $ b_n $ is positive for all $ n \geq N $ for some $ N $ and is easier to evaluate.

3. Take the limit of the ratio of the two series as $ n $ approaches infinity:

$$ \lim_{{n \to \infty}} \frac{a_n}{b_n} $$

4. If the limit is a finite positive number, then either both series converge or both series diverge. If the limit is zero or infinity, the comparison is inconclusive.

For this series, you can compare it to the series $ \sum \frac{1}{n^{3/2}} $ because $ \frac{\sqrt{n+1}}{n^{3/2}} $ is a simpler expression and easier to evaluate.

Taking the limit of the ratio:

$$ \lim_{{n \to \infty}} \frac{\sqrt{n+1}}{n^{3/2}} = \lim_{{n \to \infty}} \frac{\sqrt{n+1}}{n^{3/2}} \cdot \frac{\frac{1}{\sqrt{n}}}{\frac{1}{\sqrt{n}}} = \lim_{{n \to \infty}} \frac{\sqrt{n+1}}{n} \cdot \frac{1}{\sqrt{n}} = 0 $$

Since the limit is a finite positive number (zero), by the limit comparison test, both series $ \sum \frac{\sqrt{n+1}}{n^2 + 1} $ and $ \sum \frac{1}{n^{3/2}} $ either converge or diverge together.