\begin{eqnarray*}e^x=\sum_{n=0}^\infty\frac{x^n}{n!}&\rightarrow&\frac{a_{n+1}}{a_n}=\frac{x^{n+1}}{(n+1)!}\;\frac{n!}{x^n}=\frac{x}{n+1}\stackrel{n\to\infty}\longrightarrow0\\ &\rightarrow&\mbox{convergent for every $x$}\end{eqnarray*}
\begin{eqnarray*} \sum_{n=1}^\infty\frac{1}{n^2}&\rightarrow&\frac{a_{n+1}}{a_n}=\frac{\frac{1}{(n+1)^2}}{\frac{1}{n^2}}\rightarrow1\;\;?\\ &&\sqrt[n]{a^n}=\frac{1}{\sqrt[n]{n^2}}=\frac{1}{\left(\sqrt[n]{n}\right)^2}\rightarrow1\;\;?\\ \mbox{in fact}&&\sum_{n=1}^\infty\frac{1}{n^2}=\frac{\pi^2}{6}\;\;!\end{eqnarray*}
\[ \sum_{n=1}^\infty x^{n!}\qquad\frac{a_{n+1}}{a_n}=\frac{x^{(n+1)!}}{x^{n!}}=x^{(n+1)n!-n!}=x^{n\cdot n!}\rightarrow\begin{array}{l} \mbox{converges if $|x|\lt1$}\\ \mbox{diverges if $|x|\gt1$}\\ \mbox{diverges if $|x|=1$} \end{array} \] |
© J. Carstensen (Math for MS)