Ratio test

For nonzero terms, ifthen $\sum a_n$ converges absolutely. If the ratio has a limit greater than one, the terms do not tend to zero and the series diverges.

Suppose $a_n>0$ and, for all sufficiently large $n$, $a_{n+1}/a_n\le 1-c/n$ with fixed $c>0$. Choose an initial index $N>c$. Since $\log (1-x)\le -x$ for $0<x<1$, summing logarithmic ratios gives $\log (a_n/a_N)\le -c{\sum }_{k=N}^{n-1}1/k\le -c\log (n/N)$. Thus $a_n\le a_N{N}^c{n}^{-c}$. For $c>1$, the comparison test for series with a P-series proves convergence. This supplies a useful refinement when a limiting ratio of one makes the ordinary ratio test inconclusive.

For a positive sequence, if $a_{n+1}/a_n\to L\ge 0$, thenFor $L>0$ this follows by taking logarithms and applying Cesàro averaging; the case $L=0$ follows from an eventual geometric upper bound.

For $a_n=n!/n^n$,Thus $\sum n!/n^n$ converges and $n/(n!{)}^{1/n}\to e$.