## Related questions with answers

What is the Ratio Test inconclusive for summation n=1 to infinite (-3)^n-1/n^1/2 (that is, it fails to give a definite answer)?

Solutions

Verified$a_n = \dfrac{(-3)^{n-1}}{\sqrt{n}} \Rightarrow a_{n+1} = \dfrac{(-3)^n}{\sqrt{n+1}} \Rightarrow \lim\limits_{n \to \infty} \left|\dfrac{a_{n+1}}{a_{n}} \right| = \lim\limits_{n \to \infty} \left|\dfrac{\dfrac{(-3)^n}{\sqrt{n+1}}}{\dfrac{(-3)^{n-1}}{\sqrt{n}}} \right| = \lim\limits_{n \to \infty} \dfrac{3^n\sqrt{n}}{3^{n-1}\sqrt{n+1}} = \lim\limits_{n \to \infty} \dfrac{3\sqrt{n}}{\sqrt{n+1}} =\lim\limits_{n \to \infty} \dfrac{3\sqrt{n}}{\sqrt{n}} = 3> 1$

$\therefore$ The series diverges by the Ratio Test.

$\textbf{Ratio Test:}$

(1) If $\lim\limits_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n} \right| < 1$, then $\sum a_n$ converges (absolutely).

(2) If $\lim\limits_{n \to \infty}\left|\dfrac{a_{n+1}}{a_n} \right| > 1$, then $\sum a_n$ diverges.

(3) If $\lim\limits_{n \to \infty} \left|\dfrac{a_{n+1}}{a_n} \right| = 1$, then the Ratio Test is inconclusive.

Note: In general, Ratio Test is inconclusive (thus not applicable) if there is $\text{\underline{no}}$ variable in the exponent (i.e., there is no $n$th power).

Let's test the convergence of the series by using the Ratio Test:

$\begin{align*} \left| \frac{\frac{(-3)^{n}}{\sqrt{n+1}}}{\frac{(-3)^{n-1}}{\sqrt{n}}} \right| = \frac{\frac{3^{n}}{\sqrt{n+1}}}{\frac{3^{n-1}}{\sqrt{n}}} = 3\frac{\sqrt{n}}{\sqrt{n+1}} \longrightarrow 3 > 1 \end{align*}$

Therefore, the Ratio Test is conclusive for this series - it is divergent.

We need to determine if the **Ratio Test** is **Inconclusive** to the given series.

$\begin{aligned} \sum_{n=1}^{\infty}\frac{(-3)^{n-1}}{\sqrt{n}} \end{aligned}$

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