Bourbaki

Exercise 7, § 4

Functions of a Real Variable, V, p. 50

Let (un)(u_n) be a convergent series with terms >0> 0. a) Now prove that

lim⁡n→∞u1+2u2+⋯+nunn=0. \lim_{n \to \infty} \frac{u_1 + 2u_2 + \cdots + nu_n}{n} = 0.
b) Now prove that
∑n=1∞u1+2u2+⋯+nunn(n+1)=∑n=1∞un \sum_{n=1}^\infty \frac{u_1 + 2u_2 + \cdots + nu_n}{n(n+1)} = \sum_{n=1}^\infty u_n
(write 1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}). c) Deduce from a) and b) that one has
lim⁡n→∞n(u1u2…un)1/n=0 \lim_{n \to \infty} n(u_1 u_2 \ldots u_n)^{1/n} = 0
and
∑n=1∞(n!u1u2…un)1/nn+1<∑n=1∞un \sum_{n=1}^\infty \frac{(n! u_1 u_2 \ldots u_n)^{1/n}}{n+1} < \sum_{n=1}^\infty u_n
(apply inequality (8) of III, p. 3).

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.