Question

Convergence or divergence of a series: For each of the series that follow, determine whether the series converges or diverges. Explain the criteria you are using and why your conclusion is valid.

$\sum _{k=1}^{\infty }\frac{k}{3^k}$

Short answer

By ratio test, the sequence is convergent.

Step-by-step solution

Step 1. Given Information

The given sequence is$\sum _{k=1}^{\infty }\frac{k}{3^k}$.

Step 2. Apply the ratio test

• The ratio test states that for the sequence $\sum _{k=1}^{\infty }{a}_k$, determine the value, $p=\underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}$. If $p<1$, the series converges. If $p>1$, the series diverges. Otherwise, it is inconclusive.
• Determine the value of $p$.

$\begin{array}{rcl}\underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}& =& \underset{k\to \infty }{\lim }\frac{\left({\displaystyle \frac{k+1}{3^{k+1}}}\right)}{\left({\displaystyle \frac{k}{3^k}}\right)}\\ & =& \underset{k\to \infty }{\lim }\left(\frac{k+1}{k}\right)3^{k-\left(k+1\right)}\\ & =& \underset{k\to \infty }{\lim }\left(1+\frac{1}{k}\right)3^{-1}\\ & =& \left(1+\frac{1}{\infty }\right)3^{-1}\\ & =& 1\left(\frac{1}{9}\right)\\ & =& \frac{1}{9}\\ & <& 1\end{array}$

• Thus, by ratio test, the sequence is convergent.