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Exercise 1509Calculus › Single-variable calculus › SequencesL1Added 2 Nov 2026

The Cesàro lemma

#Cesaro mean#epsilon-N#averages

Statement

  1. Prove that if un→ℓu_n\to\ell then u1+⋯+unn→ℓ\dfrac{u_1+\dots+u_n}n\to\ell.
  2. Deduce that if un+1−un→ℓu_{n+1}-u_n\to\ell then unn→ℓ\dfrac{u_n}n\to\ell.
  3. Show that the converse of part 1 is false.
Hint

For part 1, assume ℓ=0\ell=0 (subtract ℓ\ell from every term). Split the average into the first NN terms and the others.

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Solution

1. Replace unu_n by un−ℓu_n-\ell: we may assume ℓ=0\ell=0. Let ε>0\varepsilon>0 and choose NN with ∣uk∣<ε|u_k|<\varepsilon for k>Nk>N. Then
∣u1+⋯+unn∣≤∣u1∣+⋯+∣uN∣n+n−Nnε≤Cn+ε,\left|\frac{u_1+\dots+u_n}n\right|\le\frac{|u_1|+\dots+|u_N|}n+\frac{n-N}n\varepsilon\le\frac C n+\varepsilon,
with C=∑k≤N∣uk∣C=\sum_{k\le N}|u_k| fixed. For nn large, Cn<ε\frac Cn<\varepsilon, so the average is below 2ε2\varepsilon.

2. Write un=u0+∑k=0n−1(uk+1−uk)u_n=u_0+\sum_{k=0}^{n-1}(u_{k+1}-u_k). Then
unn=u0n+1n∑k=0n−1(uk+1−uk)⟶0+ℓ\frac{u_n}n=\frac{u_0}n+\frac1n\sum_{k=0}^{n-1}(u_{k+1}-u_k)\longrightarrow0+\ell
by part 1 applied to the sequence uk+1−uku_{k+1}-u_k.

3. un=(−1)nu_n=(-1)^n does not converge, yet its averages are 00 or −1n-\frac1n, which tend to 00.

Remark. Cesàro averaging "smooths" oscillations: it is the first example of a *summation method* for divergent sequences.