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Superiority of Root Test over Ratio Test

Suppose that $\{a(j), j = 0,\dots\}$ is a sequence of strictly positive terms with a(0) = 1. Put $A(n) = a_0 + \dots + a_n$.The weak form of the ratio test says that if

\begin{displaymath}
\lim_{j\rightarrow\infty}\frac{a(j)}{a(j-1)} = R < 1\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} A(n) \in (1,\infty),\end{displaymath}

while if

\begin{displaymath}
\lim_{j\rightarrow\infty}\frac{a(j)}{a(j-1)} = R \gt 1\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} A(n) = \infty.\end{displaymath}

The weak form of the root test says that if

\begin{displaymath}
\lim_{j\rightarrow\infty}\sqrt[j]{a(j)} = S < 1\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} A(n) \in (1,\infty),\end{displaymath}

while if

\begin{displaymath}
\lim_{j\rightarrow\infty}\sqrt[j]{a(j)} = S \gt 1\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} A(n) = \infty.\end{displaymath}

Using elementary properties of logarithms these tests can be reformulated to say:

Ratio Test: If

\begin{displaymath}
\lim_{j\rightarrow\infty}\log\left(\frac{a(j)}{a(j-1)}\right) = r < 0\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} A(n) \in (1,\infty),\end{displaymath}

while if

\begin{displaymath}
\lim_{j\rightarrow\infty}\log\left(\frac{a(j)}{a(j-1)}\right) = r \gt 0\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} A(n) = \infty.\end{displaymath}

Root Test: If

\begin{displaymath}
\lim_{j\rightarrow\infty}\frac{1}{j}\log(a(j)) = s < 0\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} A(n) \in (1,\infty),\end{displaymath}

while if

\begin{displaymath}
\lim_{j\rightarrow\infty}\frac{1}{j}\log(a(j)) = s \gt 0\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} A(n) = \infty.\end{displaymath}

Since

\begin{displaymath}
a(j) = \prod_{k=1}^j \frac{a(k)}{a(k-1)}\end{displaymath}

we have

\begin{displaymath}
\frac{1}{j}\log(a(j)) = \frac{1}{j}\sum_{k=1}^j
\log\left(\frac{a(k)}{a(k-1)}\right) \end{displaymath}

so if

\begin{displaymath}
\lim_{j\rightarrow\infty}\log\left(\frac{a(j)}{a(j-1)}\right) = r\end{displaymath}

then

\begin{displaymath}
\lim_{n\rightarrow\infty} \frac{1}{j}\log(a(j)) = r.\end{displaymath}

So the weak form of the root test will decide the convergence/divergence of A(n) whenever the weak form of the ratio test does, and potentially will decide the convergence/divergence of A(n) when the weak form of the ratio test fails to give any information. The demonstration above makes it clear why: The ratio test only uses one ratio, while the root test uses all the ratios. One might say, the root test is better because it remembers it roots.



 
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Eric S Key
3/23/1999