Tannery's theorem

Let ${\displaystyle S_{n}=\sum _{k=0}^{\infty }a_{k}(n)}$ and suppose that ${\displaystyle \lim _{n\to \infty }a_{k}(n)=b_{k}}$. If ${\displaystyle |a_{k}(n)|\leq M_{k}}$ and ${\displaystyle \sum _{k=0}^{\infty }M_{k}<\infty }$, then ${\displaystyle \lim _{n\to \infty }S_{n}=\sum _{k=0}^{\infty }b_{k}}$.^[2][3]

Tannery's theorem follows directly from Lebesgue's dominated convergence theorem applied to the sequence space ${\displaystyle \ell ^{1}}$.

An elementary proof can also be given.^[3]

Tannery's theorem can be used to prove that the binomial limit and the infinite series characterizations of the exponential ${\displaystyle e^{x}}$ are equivalent. Note that

${\displaystyle \lim _{n\to \infty }\left(1+{\frac {x}{n}}\right)^{n}=\lim _{n\to \infty }\sum _{k=0}^{n}{n \choose k}{\frac {x^{k}}{n^{k}}}.}$

Define ${\displaystyle a_{k}(n)={n \choose k}{\frac {x^{k}}{n^{k}}}}$. We have that ${\displaystyle |a_{k}(n)|\leq {\frac {|x|^{k}}{k!}}}$ and that ${\displaystyle \sum _{k=0}^{\infty }{\frac {|x|^{k}}{k!}}=e^{|x|}<\infty }$, so Tannery's theorem can be applied and

${\displaystyle \lim _{n\to \infty }\sum _{k=0}^{\infty }{n \choose k}{\frac {x^{k}}{n^{k}}}=\sum _{k=0}^{\infty }\lim _{n\to \infty }{n \choose k}{\frac {x^{k}}{n^{k}}}=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}=e^{x}.}$