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Answer Key for Quiz 5 (section B)
- Start by using the ratio test:
an =
(x − 3)
n
n(n + 1)
, so an+1 =
(x − 3)
n+
(n + 1)(n + 2)
and therefore
lim n→∞
an+
an
= lim n→∞
(x − 3) n+
(n + 1)(n + 2)
n(n + 1)
(x − 3) n
= lim n→∞
(x − 3)
n
n + 2
= |x − 3 | lim n→∞
2 n
= |x − 3 |
= |x − 3 |.
So the series converges if |x − 3 | < 1, it diverges if |x − 3 | > 1, and we don’t know yet what it does if
|x − 3 | = 1. If |x − 3 | = 1 then either x − 3 = 1 or x − 3 = −1; in other words, either x = 4 or x = 2. We
know so far that the series converges if 2 < x < 4 and that the only other places where it might converge
are x = 2 and x = 4. Plugging x = 4 into the series we get
∞ ∑
n=
n
n(n + 1)
∞ ∑
n=
n(n + 1)
This converges, either because
∞ ∑
n=
n(n + 1)
∞ ∑
n=
n 2
which is finite (it equals
π 2
or because
∞ ∑
n=
n(n + 1)
∞ ∑
n=
n + 1 − n
n(n + 1)
∞ ∑
n=
n
n + 1
When x = 2 the series becomes ∞ ∑
n=
n
n(n + 1)
∞ ∑
n=
n
n(n + 1)
This converges by the alternating series test because the non-alternating part, namely
1 n(n+1) , decreases to
zero. We could also say it converges absolutely (which implies that it converges), since if we put absolute
values on the terms we would get
n=
1 n(n+1)
, which we already know converges. So this series converges at
both endpoints: the interval of convergence is 2 ≤ x ≤ 4.
I didn’t ask this, but we can find what function the series converges to by using the 1 = n + 1 − n trick:
∞ ∑
n=
(x − 3)
n
n(n + 1)
∞ ∑
n=
n + 1 − n
n(n + 1)
(x − 3)
n
∞ ∑
n=
(x − 3)
n
n
∞ ∑
n=
(x − 3)
n
n + 1
We know that
(L) − ln(1 − z) =
∞ ∑
n=
z
n
n
for − 1 ≤ z < 1
which will give us the first series above. To get the second one, rewrite (L) as
ln(1 − z)
z
∞ ∑
n=
z
n− 1
n
∞ ∑
n=
z
n
n + 1
Since the first term of this series is 1, if we subtract 1 from both sides we get
(LL) −
ln(1 − z)
z
∞ ∑
n=
z
n
n + 1
Taking z = x − 3 in (L) and (LL) we get
∞ ∑
n=
(x − 3)
n
n
∞ ∑
n=
(x − 3)
n
n + 1
= − ln (1 − (x − 3)) +
ln (1 − (x − 3))
x − 3
ln(4 − x)
x − 3
− ln(4 − x) = 1 +
x − 4
x − 3
ln(4 − x).
This holds if 2 ≤ x ≤ 4, although one needs to take a limit if x = 4.
- We have to plug into the formula
f (a) + f
′ (a)(x − a) + f
′′ (a)
(x − a)
2
′′′ (a)
(x − a)
3
with a = 7 and f (x) =
8 − x
. We can plug the 7 in right away: we need to work out
(T) f (7) + f
′ (7)(x − 7) + f
′′ (7)
(x − 7)
2
′′′ (7)
(x − 7)
3
In particular we need the first three derivatives of f (x) = (8 − x)
− 1
. Using the chain rule, we have
f
′ (x) = (−1)(8 − x)
− 2 (−1) = (8 − x)
− 2
(8 − x) 2
f
′′ (x) = (−2)(8 − x)
− 3 (−1) = 2(8 − x)
− 3
(8 − x) 3
f
′′′ (x) = 2(−3)(8 − x)
− 4 (−1) = 6(8 − x)
− 4
(8 − x) 4
Plugging in x = 7 we get f (7) = 1, f
′ (7) = 1, f
′′ (7) = 2 and f
′′′ (7) = 6. Putting these numbers in formula
(T) we get the answer
1 + (x − 7) + 2
(x − 7)
2
(x − 7)
3
= 1 + (x − 7) + (x − 7)
2
3 .
We can get the full Taylor series by realizing that
f
(n) (x) =
n!
(8 − x)n+^
and therefore f
(n) (7) =
n!
(8 − 7)n+^
= n!
Therefore the general formula
∞ ∑
n=
f
(n) (7)
(x − 7)
n
n!
becomes
∞ ∑
n=
n!
(x − 7)
n
n!
∞ ∑
n=
(x − 7)
n .
Another way to get this is to recall that
1 − r
∞ ∑
n=
r
n if − 1 < r < 1.
If we replace r by x − 7 then this becomes
8 − x
1 − (x − 7)
∞ ∑
n=
(x − 7)
n if − 1 < x − 7 < 1, which means 6 < x < 8.
