How do you use the limit comparison test on the series #sum_(n=1)^oo1/sqrt(n^3+1)# ?

Question

Elijah Abram · Accepted Answer

The Limit Comparison Test tells you that if #0 < lim_(n->oo)a_n/b_n < oo#, with #a_n>0 and b_n >0,# then #sum_(n=1)^(oo)a_n# converges if and only if #sum_(n=1)^(oo)b_n# converges. We want to find a series whose terms #b_n# compare to our #a_n = 1/(sqrt(n^3+1))#; let's choose #b_n = 1/(sqrt(n^3))=1/(n^(3/2))#. So #sum_(n=1)^(oo)b_n# converges by the Integral p-test; #p = 3/2 > 1.# #lim_(n->oo)a_n/b_n=lim_(n->oo)(sqrt(n^3))/(sqrt(n^3+1))=sqrt(lim_(n->oo)(n^3)/(n^3+1))=1# Since #0 < 1 < oo,# our series converges: 1. by the Limit Comparison Test, and 2. by the \ illustrious dansmath / ;-}

Aurora Alvarado · Answer

undefined To use the limit comparison test on the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 1}}$, we compare it with a known series whose convergence or divergence is known.

Let's denote the given series as $a_n = \frac{1}{\sqrt{n^3 + 1}}$.

Choose a series $b_n$ such that:

1. $b_n > 0$ for all $n$,
2. $\lim_{n \to \infty} \frac{a_n}{b_n}$ exists and is positive and finite.

One common choice for $b_n$ that satisfies these conditions is $b_n = \frac{1}{n^{3/2}}$.

Now, we find the limit:

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

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

Now, we can rationalize the denominator:

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

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

Since $n^{3/2}$ grows slower than $\sqrt{n^3 + 1}$ as $n \to \infty$, we can simplify the limit to:

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

Since the limit is a finite positive number, the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 1}}$ converges or diverges in the same way as the series $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$.

The series $\sum_{n=1}^{\infty} \frac{1}{n^{3/2}}$ is a $p$-series with $p = \frac{3}{2}$, and since $p > 1$, it converges.

Therefore, by the limit comparison test, the original series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n^3 + 1}}$ also converges.