Kronecker's Lemma

Theorem

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

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

exists and is finite.

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

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

Proof

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

Using Summation by Parts:

$\ds \frac 1 {b_n} \sum_{k \mathop = 1}^n b_k x_k = S_n - \frac 1 {b_n} \sum_{k \mathop = 1}^{n - 1} \paren {b_{k + 1} - b_k} S_k$

Now, pick any $\epsilon \in \R_{>0}$.

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 right hand side is:

\(\ds \)

\(\)

\(\ds S_n - \frac 1 {b_n} \sum_{k \mathop = 1}^{N - 1} \paren {b_{k + 1} - b_k} S_k - \frac 1 {b_n} \sum_{k \mathop = N}^{n - 1} \paren {b_{k + 1} - b_k} S_k\)

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

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 $\epsilon \dfrac {b_n - b_N} {b_n} \le \epsilon$.

$\blacksquare$

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Source of Name

This entry was named for Leopold Kronecker.

Sources

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• 2014: Christopher Clapham and James Nicholson: The Concise Oxford Dictionary of Mathematics (5th ed.) ... (previous) ... (next): Kronecker's Lemma