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I had the following homework question where I had to determine if the following expression converges by using the root test.

$$\sum_{n = 1} \frac{n}{3^n}$$

This is what I tried:

$$\begin{align} & \sum_{n = 1} \frac{n}{3^n} = \\ & \sum_{n = 1} \left( \frac{\sqrt[n]{n}}{3} \right)^n \rightarrow \\ & \lim_{n \to \infty} \sqrt[n]{\left( \frac{\sqrt[n]{n}}{3} \right)^n} = \\ & \lim_{n \to \infty} \frac{\sqrt[n]{n}}{3} = \\ & \infty \end{align}$$

Since the limit is greater then one, I concluded that the expression diverges.

However, I tried using the ratio test to confirm my answer, and obtained the following:

$$\begin{align} & \sum_{n = 1} \frac{n}{3^n} \rightarrow \\ & \lim_{n \to \infty} \dfrac{\frac{n + 1}{3^{n + 1}}}{\frac{n}{3^n}} = \\ & \lim_{n \to \infty} \frac{n + 1}{3 \cdot 3^n} \cdot \frac{3^n}{n} = \\ & \lim_{n \to \infty} \frac{n + 1}{3n} = \\ & \frac{1}{3} \end{align}$$

Since the limit is less then one, according to the ratio test, the sequence converges.

What gives? Have I misunderstood how to apply the root and ratio test, or is there an error in my math somewhere?

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1 Answer 1

up vote 5 down vote accepted

The root test gives you $$\lim_{n\rightarrow \infty}\frac{n^{1/n}}{3}$$ where $n^{1/n}\rightarrow 1$ by L.Hospital Rule. So this is definitely less than 1.

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