Ivan Shishkin, Rye (1878)

Problems/Sequence and seriesReviewed

Sequence defined by its sums of powers

by Sequoia·
52
Difficulty scaleÉchelle de difficulté

This score reflects both the level of the required concepts and the difficulty of the solution.Ce score tient compte à la fois du niveau des notions nécessaires et de la difficulté de la résolution.

  1. 110First steps / middle schoolPremiers pas / collège
  2. 1125Beginner / high schoolDébutant / lycée
  3. 2650Intermediate / undergraduateIntermédiaire / licence
  4. 5170Advanced / graduateAvancé / master
  5. 7190Expert / specializedExpert / spécialisé
  6. 91100Research levelNiveau recherche
These levels are approximate guides.Ces niveaux sont des repères approximatifs.
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Is there a sequence (an)n(a_n)_n of real numbers such that
kN,n=0+ank=1k2?\forall k\in\N^*, \sum_{n=0}^{+\infty}a_n^k=\frac{1}{k^2} ?

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Solution by anoFR

Discussions1 useful vote

Non. Raisonnement par l’absurde.
Pour k=2k=2, an2=14    n,an214    a12\sum a_n^2 = \frac{1}{4} \implies \forall n, a_n^2 \le \frac{1}{4} \implies ||a||_\infty \le \frac{1}{2}.
Pour k=2mk=2m, la norme 2m\ell^{2m} de la suite vaut :
a2m=(n=0+an2m)12m=(14m2)12m||a||_{2m} = \left( \sum_{n=0}^{+\infty} a_n^{2m} \right)^{\frac{1}{2m}} = \left( \frac{1}{4m^2} \right)^{\frac{1}{2m}}Par passage à la limite : limm+a2m=limm+exp(ln(4m2)2m)=1\lim_{m \to +\infty} ||a||_{2m} = \lim_{m \to +\infty} \exp\left(-\frac{\ln(4m^2)}{2m}\right) = 1.
Or, dans l’espace p\ell^p, la limite des normes est le supremum : limm+a2m=a\lim_{m \to +\infty} ||a||_{2m} = ||a||_\infty.
On obtient a=1||a||_\infty = 1, ce qui contredit a12||a||_\infty \le \frac{1}{2}. La suite n’existe pas.

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