Kronecker's lemma/Proof

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For the statement of the lemma, see Kronecker's lemma.

Let Sk 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

Pick any ε > 0. Now choose N so that Sk is ε-close to s for k > N. This can be done as the sequence Sk converges to s. Then the right hand side is:

S_n - \frac1{b_n}\sum_{k=1}^{N-1}(b_{k+1} - b_k)S_k - \frac1{b_n}\sum_{k=N}^{n-1}(b_{k+1} - b_k)S_k
= S_n - \frac1{b_n}\sum_{k=1}^{N-1}(b_{k+1} - b_k)S_k - \frac1{b_n}\sum_{k=N}^{n-1}(b_{k+1} - b_k)s - \frac1{b_n}\sum_{k=N}^{n-1}(b_{k+1} - b_k)(S_k - s)
= S_n - \frac1{b_n}\sum_{k=1}^{N-1}(b_{k+1} - b_k)S_k - \frac{b_n-b_N}{b_n}s - \frac1{b_n}\sum_{k=N}^{n-1}(b_{k+1} - b_k)(S_k - s)

Now, let n go to infinity. 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 \leq \epsilon.