Question

Use the Limit Comparison Test to determine whether the series converges.

Answer 1

Answer:

The infinite series $\displaystyle \sum\limits_{k = 1}^{\infty} \frac{8/k}{\sqrt{k + 7}}$ indeed converges.

Step-by-step explanation:

The limit comparison test for infinite series of positive terms compares the convergence of an infinite sequence (where all terms are greater than zero) to that of a similar-looking and better-known sequence (for example, a power series.)

For example, assume that it is known whether $\displaystyle \sum\limits_{k = 1}^{\infty} b_k$ converges or not. Compute the following limit to study whether $\displaystyle \sum\limits_{k = 1}^{\infty} a_k$ converges:

$\displaystyle \lim\limits_{k \to \infty} \frac{a_k}{b_k}\; \begin{tabular}{l}\\ \leftarrow Series whose convergence is known\end{tabular}$.

• If that limit is a finite positive number, then the convergence of the these two series are supposed to be the same.
• If that limit is equal to zero while $a_k$ converges, then $b_k$ is supposed to converge, as well.
• If that limit approaches infinity while $a_k$ does not converge, then $b_k$ won't converge, either.

Let $a_k$ denote each term of this infinite Rewrite the infinite sequence in this question:

$\begin{aligned}a_k &= \frac{8/k}{\sqrt{k + 7}}\\ &= \frac{8}{k\cdot \sqrt{k + 7}} = \frac{8}{\sqrt{k^2\, (k + 7)}} = \frac{8}{\sqrt{k^3 + 7\, k^2}} \end{aligned}$.

Compare that to the power series $\displaystyle \sum\limits_{k = 1}^{\infty} b_k$ where $\displaystyle b_k = \frac{1}{\sqrt{k^3}} = \frac{1}{k^{3/2}} = k^{-3/2}$. Note that this

Verify that all terms of $a_k$ are indeed greater than zero. Apply the limit comparison test:

$\begin{aligned}& \lim\limits_{k \to \infty} \frac{a_k}{b_k}\; \begin{tabular}{l}\\ \leftarrow Series whose convergence is known\end{tabular}\\ &= \lim\limits_{k \to \infty} \frac{\displaystyle \frac{8}{\sqrt{k^3 + 7\, k^2}}}{\displaystyle \frac{1}{{\sqrt{k^3}}}}\\ &= 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{k^3}{k^3 + 7\, k^2}}\right) = 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{1}{\displaystyle 1 + (7/k)}}\right)\end{aligned}$.

Note, that both the square root function and fractions are continuous over all real numbers. Therefore, it is possible to move the limit inside these two functions. That is:

$\begin{aligned}& \lim\limits_{k \to \infty} \frac{a_k}{b_k}\\ &= \cdots \\ &= 8\left(\lim\limits_{k \to \infty} \sqrt{\frac{1}{\displaystyle 1 + (7/k)}}\right)\\ &= 8\left(\sqrt{\frac{1}{\displaystyle 1 + \lim\limits_{k \to \infty} (7/k)}}\right) \\ &= 8\left(\sqrt{\frac{1}{1 + 0}}\right) \\ &= 8 \end{aligned}$.

Because the limit of this ratio is a finite positive number, it can be concluded that the convergence of $\displaystyle a_k &= \frac{8/k}{\sqrt{k + 7}}$ and $\displaystyle b_k = \frac{1}{\sqrt{k^3}}$ are the same. Because the power series $\displaystyle \sum\limits_{k = 1}^{\infty} b_k$ converges, (by the limit comparison test) the infinite series $\displaystyle \sum\limits_{k = 1}^{\infty} a_k$ should also converge.