72
II - Convergence: Discrete variables
Example 4. The sequence Un = (_l)n + l/n does not converge; its terms of
even order tend to 1, its terms of odd order to -1.
We note in this connection that if a sequence (un) converges to a limit u,
then every subsequence that one can extract from it will converge, also to u.
Such a subsequence is obtained by choosing an increasing sequence of integers
PI, P2, etc. and omitting the terms of the initial sequence except for those
corresponding to this choice. Convergence then follows from the obvious fact
that Pn ~ n for all n.
For example, the sequence with general term 1/n 3 tends to 0, since it is
a subsequence of the sequence of example 2.
Example 5. If q is a complex number, then
(5.3)
limqn = 0 if Iql < 1.
We need to show that, for any r > 0, one has Iqn I < r for n large. By replacing
q by Iql, one reduces to the case where q ~ 0, and even to q > 0, since the
case where q = 0 is trivial. As we are now assuming that q < 1, we have
1 = q + t with t > o. All the terms in the binomial formula
are positive, whence (n + l)qnt < 1, or
o < qn < l/t(n + 1).
The right hand side clearly tends to 0, so qn does too.
For q = 1, the limit is clearly 1. For all other possible values of q the
sequence is divergent. For suppose that lim qn = u exists for some number
q E C. The sequence with general term qn+l also converges to u, since it is
a subsequence. But qn+l = q.qn, and it follows from the definition that, for
every convergent sequence,
lim Un = u implies limq.un = q.u
for any q E
a relation which is satisfied for large n since r /Iql > O. Returning to the
sequence considered, we must have qu = u, Le. either q = 1, the trivial case,
or u = 0, which cannot be the case for Iql ~ 1 since then Iqnl ~ 1 for all n.
We therefore have divergence except for the cases Iql < 1 and q = 1.
Example 6. Let us show that
(5.4)
limzn/n! = limz[n] = 0
II - Convergence: Discrete variables
Example 4. The sequence Un = (_l)n + l/n does not converge; its terms of
even order tend to 1, its terms of odd order to -1.
We note in this connection that if a sequence (un) converges to a limit u,
then every subsequence that one can extract from it will converge, also to u.
Such a subsequence is obtained by choosing an increasing sequence of integers
PI, P2, etc. and omitting the terms of the initial sequence except for those
corresponding to this choice. Convergence then follows from the obvious fact
that Pn ~ n for all n.
For example, the sequence with general term 1/n 3 tends to 0, since it is
a subsequence of the sequence of example 2.
Example 5. If q is a complex number, then
(5.3)
limqn = 0 if Iql < 1.
We need to show that, for any r > 0, one has Iqn I < r for n large. By replacing
q by Iql, one reduces to the case where q ~ 0, and even to q > 0, since the
case where q = 0 is trivial. As we are now assuming that q < 1, we have
1 = q + t with t > o. All the terms in the binomial formula
are positive, whence (n + l)qnt < 1, or
o < qn < l/t(n + 1).
The right hand side clearly tends to 0, so qn does too.
For q = 1, the limit is clearly 1. For all other possible values of q the
sequence is divergent. For suppose that lim qn = u exists for some number
q E C. The sequence with general term qn+l also converges to u, since it is
a subsequence. But qn+l = q.qn, and it follows from the definition that, for
every convergent sequence,
lim Un = u implies limq.un = q.u
for any q E
a relation which is satisfied for large n since r /Iql > O. Returning to the
sequence considered, we must have qu = u, Le. either q = 1, the trivial case,
or u = 0, which cannot be the case for Iql ~ 1 since then Iqnl ~ 1 for all n.
We therefore have divergence except for the cases Iql < 1 and q = 1.
Example 6. Let us show that
(5.4)
limzn/n! = limz[n] = 0