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Answer Key for Quiz 5
- If an =
(2n)!
n! (n + 1)!
x
n , then
an+1 =
(2(n + 1))!
(n + 1)! (n + 2)!
x
n+
(2n + 2)!
(n + 1)! (n + 2)!
x
n+
and we want to look at
lim n→∞
an+
an
= lim n→∞
(2n+2)! (n+1)! (n+2)! x
n+
(2n)! n! (n+1)! x n
= lim n→∞
(2n + 2)!
(2n)!
n! (n + 1)!
(n + 1)! (n + 2)!
x n+
x n
= lim n→∞
(2n + 2)!
(2n)!
n!
(n + 2)!
|x|.
Since
(m + 2)! = (m + 2)(m + 1)(m)(m − 1) · · · 3 · 2 · 1 = (m + 2)(m + 1) × m!,
this simplifies to
lim n→∞
an+
an
= lim n→∞
(2n + 2)(2n + 1)
(n + 2)(n + 1)
|x| = lim n→∞
2 n + 1
n + 2
2 |x|.
Either by L’Hˆopital’s rule or by dividing all terms by n we have lim n→∞
2 n + 1
n + 2
= 2, so
lim n→∞
an+
an
= 2 · 2 |x| = 4|x|.
Therefore the series converges absolutely if 4|x| < 1, diverges if 4|x| > 1, and we don’t know yet what it does
if 4|x| = 1. In other words, the series converges on an interval centered at x = 0 with endpoints ±
1 4 , so the
radius of convergence is
1 4
Checking the endpoints isn’t easy. To check x =
1 4
we need to look at the series whose general term is
(2n)!
n! (n + 1)!
n
1 · 2 · 3 · · · (2n)
[(1 · 2 · 3 · · · n) 2 n ]
2
n + 1
1 · 2 · 3 · · · (2n)
[2 · 4 · 6 · · · (2n)]
2
n + 1
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
n + 1
Now
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
2 · 4 · 6 · · · (2n)
3 · 5 · 7 · · · (2n + 1)
and if we multiply both sides of this by the quantity on the left side we get
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
2 · 4 · 6 · · · (2n)
3 · 5 · 7 · · · (2n + 1)
2 n + 1
So
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
2 n + 1
and therefore
(2n)!
n! (n + 1)!
n
1 · 3 · 5 · · · (2n − 1)
2 · 4 · 6 · · · (2n)
n + 1
n + 1
2 n + 1
This tells us that
∞ ∑
n=
(2n)!
n! (n + 1)!
n
∞ ∑
n=
n + 1
2 n + 1
∞ ∑
n=
(n + 1)
3 2
∞ ∑
n=
n
3 2
which converges since it is a p-series with p =
3 2
. Therefore the given power series converges when x =
1 4
When x = −
1 4 it becomes ∞ ∑
n=
(2n)!
n! (n + 1)!
)n
and this converges absolutely, since if we put absolute values on it we get
∞ ∑
n=
(2n)!
n! (n + 1)!
n
which converges as we just saw. Hence the interval of convergence is −
≤ x ≤
, and we actually have
∞ ∑
n=
(2n)!
n! (n + 1)!
x
n
1 − 4 x
2 x
if −
≤ x ≤
One can derive this formula by integrating the binomial series with p = −
1 2
and x replaced by − 4 x.
- The first three terms of the Taylor series of a generic function f (x) at x = 1 are
f (1) + f
′ (1) (x − 1) + f
′′ (1)
(x − 1) 2
If f (x) = x
x , then we are given f
′ (x) = x
x (1 + ln x), so we only need to calculate f ′′ (x), which we can do
by the product rule:
f
′′ (x) = x
x d
dx
(1 + ln x) + (1 + ln x)
d
dx
x
x
= x
x
x
x (1 + ln x)
= x
x− 1
x (1 + ln x)
2 .
Since ln 1 = 0, this gives
f (1) = 1
1 = 1 and f
′ (1) = 1
1 (1 + ln 1) = 1 + 0 = 1 and f
′′ (1) = 1
1 − 1
1 (1 + ln 1)
2 = 1 + 1
2 = 2,
so the first three terms of the Taylor series are
1 + (x − 1) + 2
(x − 1)
2
This implies that
x
x ≈ x + (x − 1)
2 if x is near 1,
so, for example, (.98)
. 98 ≈ .98 + (−.02)
2 = .9804, which is pretty close.
To check the formula for f ′ (x) we can start with
ln {f (x)} = ln (x
x ) = x ln x
and then take the derivative on both sides:
f (x)
f
′ (x) = x ·
x
so
f
′ (x) = f (x) (1 + ln x) = x
x (1 + ln x).
