Kronecker's Lemma

Theorem
Let $\left \langle {x_n} \right \rangle$ be an infinite sequence of real numbers such that:
 * $\displaystyle \sum_{n \mathop = 1}^\infty x_n = s$

exists and is finite.

Then for $0 < b_1 \le b_2 \le b_3 \le \ldots$ and $b_n \to \infty$:
 * $\displaystyle \lim_{n \to \infty} \frac 1 {b_n} \sum_{k \mathop = 1}^n b_k x_k = 0$

Proof
Let $S_k$ denote the partial sums of the $x$s.

Using Summation by Parts:


 * $\displaystyle\frac 1 {b_n} \sum_{k \mathop = 1}^n b_k x_k = S_n - \frac 1 {b_n}\sum_{k \mathop = 1}^{n - 1} \left({b_{k + 1} - b_k}\right) S_k$

Now, pick any $\epsilon > 0$.

Choose $N$ such that $S_k$ is $\epsilon$-close to $s$ for $k > N$.

This can be done, as the sequence $S_k$ converges to $s$.

Then the is:

Now, let $n \to \infty$.

The first term goes to $s$, which cancels with the third term.

The second term goes to zero (as the sum is a fixed value).

Since the $b$ sequence is increasing, the last term is bounded by $\epsilon \left({b_n - b_N}\right) / b_n \le \epsilon$.