How do you apply the ratio test to determine if #Sigma 1/sqrtn# from #n=[1,oo)# is convergent to divergent?

Question

Scarlett Allison · Accepted Answer

The series diverges, but this Ratio Test cannot determine this.

The Ratio Test for an infinite series says you can take an infinite series #sum_{n=1}^{oo} a_n# and possibly determine whether it converges or diverges by finding the following limit: #L = lim_{n->oo} |a_{n+1}/a_n| # Based on the results of that limit #L#, you can make these conclusions: For this problem, #a_n = 1/sqrt(n)#. Thus: #L = lim_{n->oo} |(1/sqrt(n+1))/(1/sqrt(n))| = lim_{n->oo} |sqrt(n)/sqrt(n+1)|# From here, take you choice on how to proceed. If you know L'Hopital's Rule, that is an option. Otherwise, there are other ways to evaluate this limit. (For instance, observe that for large values of #n#, the value of #n# dominates the value of 1, meaning you can consider the denominator to effectively be #sqrt(n)# and thus the limit is a limit of 1.) The problem is this limit comes out to 1, meaning the Ratio Test will not tell you the convergence or divergence of this series. You must use another test, such as the Integral Test or the Comparison Test (shown here ). In any case, the series diverges.

Evelyn Agard · Answer

undefined To apply the ratio test to the series Σ(1/√n) from n=1 to infinity, you need to compute the limit of the ratio of successive terms as n approaches infinity.

First, find the general term of the series, which is $ a_n = \frac{1}{\sqrt{n}} $.

Then, compute the ratio of successive terms:
$$ \frac{a_{n+1}}{a_n} = \frac{\frac{1}{\sqrt{n+1}}}{\frac{1}{\sqrt{n}}} $$

Simplify the ratio:
$$ \frac{\sqrt{n}}{\sqrt{n+1}} $$

To evaluate the limit as n approaches infinity:
$$ \lim_{n \to \infty} \frac{\sqrt{n}}{\sqrt{n+1}} = 1 $$

Since the limit is equal to 1, the ratio test is inconclusive. Therefore, you cannot determine convergence or divergence solely using the ratio test for this series.