Kronecker's Lemma

Theorem

Let $\left \langle {x_n} \right \rangle$ be an infinite sequence of real numbers such that:

$\displaystyle \sum_{n=1}^\infty x_n = s$

exists and is finite.

Then for $0< b_1 \le b_2 \le b_3 \le \ldots$ and $b_n \to \infty$:

$\displaystyle\lim_{n \to \infty} \frac 1 {b_n} \sum_{k=1}^n b_k x_k = 0$

Proof

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

Using Summation by Parts:

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

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:

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

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

$\blacksquare$

Source of Name

This entry was named for Leopold Kronecker.