## Related questions with answers

Use the ratio test to determine whether the given series converges or diverges. $\sum_{k=1}^{\infty} \frac{2^{k}}{k^{3}}$

Solution

VerifiedUse the ratio test to determine whether this infinite series converges or diverges.

$\begin{align*} L &= \lim\limits_{n \to \infty}\left(\dfrac{a_{n+1}}{a_n}\right)\\ &= \lim\limits_{n \to \infty}\left[\dfrac{\dfrac{2n+1}{\left(n+1\right)^3}}{\dfrac{2^n}{n^3}}\right]\\ &= \lim\limits_{n \to \infty}\left[\dfrac{2^{n+1}}{\left(n+1\right)^3}\cdot\dfrac{n^3}{2^n}\right]\\ &= 2\lim\limits_{n \to \infty}\left[\dfrac{n^3}{\left(n+1\right)^3}\right]\\ &=2\lim\limits_{n \to \infty}\left(\dfrac{n}{n+1}\right)^3\\ &=2\left[\lim\limits_{n \to \infty}\left(\dfrac{n}{n+1}\right)\right]^3\\ &=2\left[\lim\limits_{n \to \infty}\left(\dfrac{\dfrac{n}{n}}{\dfrac{n}{n}+\dfrac{1}{n}}\right)\right]^3\\ &=2\left[\lim\limits_{n \to \infty}\left(\dfrac{1}{1+\dfrac{1}{n}}\right)\right]^3\\ &=2\left(\dfrac{1}{1+0}\right)^3\\ &=2\left(1\right)^3\\ &=2 \end{align*}$

Since $\left|L\right|>1$, then this series diverges.

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