Bourbaki

Exercise 7, § 4

Functions of a Real Variable, V, 262

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

a) Show that

lim⁡n→∞u1+2u2+⋯+nunn=0. \lim_{n \to \infty} \frac{u_1 + 2u_2 + \cdots + nu_n}{n} = 0.

b) Show 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

lim⁡n→∞n(u1u2…un)1/n=0 \lim_{n \to \infty} n(u_1 u_2 \ldots u_n)^{1/n} = 0
and that
∑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 the inequality (8) of III, p. 93).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.