Math 3210-3
HW 24
Solutions
Convergence of Infinite Series
1. Find the sum of each series:
∞ n
X
1
(a)
2
n=3
n
n
∞
∞ n
X
X
1
1 1
1
1X 1
1 1
=
=
=
1 = 4
2
8
2
8
2
8
1
−
2
n=3
n=0
∞
n
X
2
(b)
−
3
n=1
n X
n
∞ ∞ X
−2
2
2
1
−1=
−
−
=
−1=
−2
3
3
5
1− 3
n=0
n=1
(c)
∞
X
1
n(n + 1)(n + 2)
n=1
1
1
Notice that
=
n(n + 1)(n + 2)
2
have:
sn
So
2
n(n + 1)(n + 2)
1
=
2
1
1
. So we
−
n(n + 1) (n + 1)(n + 2)
n
X
1
i(i
+
1)(i
+ 2)
i=1
X1
1
1
−
=
2 i(i + 1) (i + 1)(i + 2)
1
1
1
1
1
1
1
=
+
+ ··· +
−
−
−
2
1·2 2·3
2·3 3·4
n(n + 1) (n + 1)(n + 2)
1 1
1
=
−
2 2 (n + 1)(n + 2)
=
∞
X
1
1
= lim sn = .
n(n + 1)(n + 2) n→∞
4
n=1
∞
X
1
√
√ converges. Justify your answer.
n
+
1+ n
n=1
√
√
√
√
√
n+1− n
n+1− n √
1
1
√
√
√
Notice that
= n + 1 − n. We have
√ =
√ ·
√ =
1
n+1+ n
n+1+ n
n+1− n
2. Determine whether or not the series
sn
=
n
X
√
√
i+1− i
i=1
=
=
√
√
√
√
√
√
√
√
√
√
( 2 − 1) + ( 3 − 2) + ( 4 − 3) + · · · + ( n − n − 1) + ( n + 1 − n)
√
−1 + n + 1
∞
X
1
√
√ = lim sn = ∞, so the series diverges.
n
+
1 + n n→∞
n=1
P
P
3. Prove that if
|an | converges and (bn ) is a bounded sequence, then
an bn converges.
Thus
Proof: Since (bn ) is bounded, there is some B ∈ R such that |bn | < B for all n. Let ǫ > 0. Then there
is some N such that if n ≥ m > N , then ||am | + |am+1 | + · · · + |an || < Bǫ . Thus |am bm + am+1 bm+1 +
· · · + an bn | ≤ |am |B + |am+1P
|B + · · · + |an |B < B[|am | + · · · + |an |] ≤ B||am | + · · · + |an || < B Bǫ = ǫ.
Therefore by Theorem 107,
an bn converges.
˜