Tannery's theorem

Remarks

Given $\varepsilon > 0$, choose $K$ with $\sum_{k > K} M_k < \varepsilon$. The tails beyond $K$ of both series are then smaller than $\varepsilon$, since $\abs{a_k} \le M_k$ too. The finitely many terms with $k \le K$ converge, so the difference of the two sums is less than $3\varepsilon$ for large $n$. Example: for real $x$ put $a_k(n) = \binom{n}{k} x^k / n^k$ for $k \le n$ and $a_k(n) = 0$ for $k > n$. Then $\sum_k a_k(n) = (1 + x/n)^n$, $a_k(n) \to x^k/k!$, and $\abs{a_k(n)} \le \abs{x}^k / k!$. So $(1 + x/n)^n \to \sum_k x^k / k! = e^x$. Without the bound of clause 2 the conclusion fails: $a_k(n) = 1$ for $k = n$ and $0$ otherwise gives sums equal to $1$ but limits $a_k = 0$.

Going deeper

For a student

Tannery's theorem is the dominated convergence theorem for the counting measure on $\N$, but it predates measure theory and has a short direct proof. A typical use is to pass from finite products or binomial expansions to their limits, as in the proof that $(1 + x/n)^n \to e^x$. Compare the Weierstrass M-test, which uses the same kind of bound to get uniform convergence of a series of functions.