Analysis Homework #7
Solutions
1. Define a sequence {an } by letting a1 = 2 and
an+1 = 4 −
1
an
for each n ≥ 1.
Show that 2 ≤ an ≤ an+1 ≤ 4 for each n ≥ 1 and find the limit of this sequence.
• Since the first two terms are a1 = 2 and a2 = 4 − 1/2 = 7/2, the statement
2 ≤ an ≤ an+1 ≤ 4
does hold when n = 1. Suppose that it holds for some n, in which case
1
1
1
1
≥
≥
≥
2
an
an+1
4
=⇒
=⇒
1
1
1
1
≤4−
≤4−
≤4−
2
an
an+1
4
2 ≤ an+1 ≤ an+2 ≤ 4.
4−
Thus, the statement holds for n + 1 as well, so it actually holds for all n ∈ N. This
shows that the given sequence is monotonic and bounded, hence also convergent; let
us denote its limit by L. Then the definition of the sequence gives
an+1 = 4 −
1
an
=⇒
L=4−
1
L
=⇒
L2 − 4L + 1 = 0,
so we can solve this quadratic equation to get
√
√
√
4 ± 16 − 4
4±2 3
L=
=
= 2 ± 3.
2
2
Since an ≥ 2 for all n, however, we must have L ≥ 2 and this implies L = 2 +
√
3.
2. Suppose {an } is a sequence of non-negative terms. Show that its sequence of partial
sums sn = a1 + a2 + . . . + an converges if and only if sn is bounded for all n.
• If sn converges, then sn is bounded by Theorem 5.7 in your notes.
• If sn is bounded, on the other hand, then sn is monotonic and bounded because
sn+1 = a1 + a2 + . . . + an + an+1 = sn + an+1 ≥ sn .
Being monotonic and bounded, sn must converge by Theorem 5.8 in your notes.
3. Define a sequence {an } by letting a1 = 1 and
an+1 =
na2n
n+1
for each n ≥ 1.
Show that 0 ≤ an+1 ≤ an ≤ 1 for each n ≥ 1 and find the limit of this sequence.
2
nan
• It is clear that an+1 = n+1
≥ 0 for each n. Moreover, a1 ≤ 1 and if we assume that
an ≤ 1 for some n, then we immediately get
na2n
≤ a2n ≤ 1
n+1
an+1 =
as well. This implies that an ≤ 1 for all n and it also implies that
an+1 =
na2n
≤ a2n ≤ an
n+1
for all n. Thus, the given sequence is monotonic and bounded, hence also convergent.
If we denote its limit by L, then we must have
an+1 =
na2n
n+1
=⇒
nL2
= L2
n→∞ n + 1
L = lim
=⇒
L = 0, 1.
Since {an } is decreasing with a2 = 1/2, however, we have L ≤ 1/2 and so L = 0.
∑
∑∞
1
4. Suppose the series ∞
n=1 an converges. Show that the series
n=1 1+an diverges.
∑
• Since ∞
n=1 an converges, we have limn→∞ an = 0 by the nth term test, so
lim
n→∞
1
= 1.
1 + an
Using the nth term test once again, we conclude that
∑∞
1
n=1 1+an
diverges.