Kronecker's lemma

The lemma

If ${\displaystyle (x_n{)}_{n=1}^{\mathrm{\infty }}}$ is an infinite sequence of real numbers such that

${\displaystyle \sum _{m=1}^{\mathrm{\infty }}{x}_m=s}$

exists and is finite, then we have for all ${\displaystyle 0<b_1\le b_2\le b_3\le \dots }$ and ${\displaystyle b_n\to \mathrm{\infty }}$ that

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{b_n}\sum _{k=1}^n{b}_k{x}_k=0.}$

Proof

Let ${\displaystyle S_k}$ denote the partial sums of the x's. Using summation by parts,

${\displaystyle \frac{1}{b_n}\sum _{k=1}^n{b}_k{x}_k=S_n-\frac{1}{b_n}\sum _{k=1}^{n-1}(b_{k+1}-b_k)S_k}$

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

${\displaystyle S_n-\frac{1}{b_n}\sum _{k=1}^{N-1}(b_{k+1}-b_k)S_k-\frac{1}{b_n}\sum _{k=N}^{n-1}(b_{k+1}-b_k)S_k}$

${\displaystyle =S_n-\frac{1}{b_n}\sum _{k=1}^{N-1}(b_{k+1}-b_k)S_k-\frac{1}{b_n}\sum _{k=N}^{n-1}(b_{k+1}-b_k)s-\frac{1}{b_n}\sum _{k=N}^{n-1}(b_{k+1}-b_k)(S_k-s)}$

${\displaystyle =S_n-\frac{1}{b_n}\sum _{k=1}^{N-1}(b_{k+1}-b_k)S_k-\frac{b_n-b_N}{b_n}s-\frac{1}{b_n}\sum _{k=N}^{n-1}(b_{k+1}-b_k)(S_k-s).}$

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 ${\displaystyle ϵ(b_n-b_N)/b_n\le ϵ}$.