SUGGESTIONS FOR ANALYSIS I - SERIES 6
Exercise 20
• By hypothesis we know that limn−→+∞ an = a. We want to prove that
limn−→+∞ sn = a as well. By definition, we have to check that for every
> 0 there exists n ∈ N such that for all n ≥ n the inequality |sn − a| < is fulfilled. Now, consider an arbitrary > 0. Since limn−→+∞ an = a by
definition of limit we know that there exists n ∈ N such that for n ≥ n
the inequality |an − a| < .
So that, for n ≥ n consider the quantity |sn − a|. We have
n
n
n
1 X
1X
1 X
|ak − a| +
ak − a ≤
|ak − a|
|sn − a| = n
n
n
k=1
k=1
k=n +1
and then it is enough to show that the two summands on the right side of
the previous inequality have limit 0.
• Consider the sequence an = 2 + (−1)n and try to write a formula for sn .
Hence, prove that sn converges by computing its limit.
Exercise 21
• First of all we notice that xn ≥ 1 for all n ∈ N. Then, to prove the inductive
step use the fact that xn ≥ 1 and the suggestion g = 1 + g1 to show the
following inequality
|xn − g|
|xn − g|
1
1
≤
|xn+1 − g| = |1 +
−1− |=
xn
g
xn g
g
and conclude the thesis with the inductive hypothesis. √
• Set x0 = 1 and define by recurrence a sequence xn+1 = 1 + xn . By induction we can prove that this sequence is positive, bounded from above and
monotone increasing. So that, it admits a limit l. By continuity arguments,
this limit will satisfy the fixed point equation
√
l = 1+l
√
and then we conclude that l = g or l = 1−2 5 . Since the sequence xn is
positive we find that l = g.
• In a similar way to what we did in points (a) and (b) define recurrence
sequences xn and yn by setting x0 = 1, xn+1 = b + xan and similarly y0 = 1,
√
yn+1 = a + b + xn . Use convergence arguments to show that these two
sequences admit limit. Then find these limits with the fixed point argument.
Exercise 22
• This is a classical result known as monotone sequence theorem. You can
look in the literature to find a detailed proof.
• Let us assume that an is monotone increasing, that is an+1 ≥ an for all
n ∈ N. We have to show that for n ∈ N we have sn+1 ≥ sn . If we develop
the computation we find
a1 + · · · + an+1
a1 + · · · + an
sn+1 =
and sn =
n+1
n
so that
a1 + · · · + an+1
a1 + · · · + an
nan+1 − a1 − · · · − an
sn+1 − sn =
−
=
n+1
n
n(n + 1)
1
2
SUGGESTIONS FOR ANALYSIS I - SERIES 6
and then it suffices to use the monotony of an to conclude that
nan+1 − a1 − · · · − an
≥ 0.
n(n + 1)
If the sequence an is monotone decreasing the proof is similar.
Exercise 23
• Implication =⇒ Let us assume that the sequence an is convergent and let
a be its limit. To prove that the sequence is bounded, let fix = 1. By
definition of limit, there exists n1 ∈ N such that for n ≥ n1 |an − a| < 1 so
that −1 + a < an < 1 + a. Now, let us consider
r = max {|an − a| | 1 ≤ n ≤ n1 }
and then δ = max{r, δ} so finally the all sequence an is contained in the
interval (a − δ, a + δ) which tell us the sequence is bounded. To prove that
the sequence has only one point of accumulation, work by contradiction
and use the fact that the sequence is convergent to a.
Implication ⇐= Let us assume our sequence bounded and let a be its unique
point of accumulation. It suffices to check that an converges to a. Let us
assume, by contradiction, that our sequence does not converge to a. Then,
there exists > 0 such that for all n ∈ N there is kn ∈ N with
(1)
|akn − a| > .
Now, consider the subsequence akn . Since our original sequence is bounded,
so does this subsequence. In particular, by the Bolzano-Weierstrass theorem
there is a convergent subsequence aknh with limit point l. By definition of
limit point, it follows that l must be a point of accumulation of an . Since
an has only one point of accumulation it follows that l = a. So that, for h
large enough we find |aknh − a| ≤ contradicting the inequality (1) which
holds for the subsequence aknh as well.
• For point (b) and (c) you have to work carefully with the definition of point
of accumulation.
• Look in the literature for some fractal constructions.