Kronecker's Lemma

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

exists and is finite.

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

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

Using Summation by Parts:


 * $$\frac1 {b_n} \sum_{k=1}^n b_k x_k = S_n - \frac1{b_n}\sum_{k=1}^{n-1}(b_{k+1} - b_k)S_k$$

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 RHS 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 $$\epsilon (b_n - b_N)/b_n \le \epsilon$$.