Theory
There are only three possibilities for the convergence of a power series centered at a a a
It converges only at x = a x=a x = a R = 0 R=0 R = 0
It converges for all real x x x R = ∞ R=\infty R = ∞
There exists R>0 such that the series converges absolutely for |x-a|<R and diverges for |x-a|>R .
The set of all x x x interval of convergence .
Exercises
Exercise 1
Find all x x x
∑ n = 0 ∞ x n n ! . \sum_{n=0}^{\infty} \frac{x^n}{n!}. n = 0 ∑ ∞ n ! x n . Show solution ↓ Hide solution ↑ Solution
Apply the Ratio Test. Let a n = x n n ! a_n=\frac{x^n}{n!} a n = n ! x n
∣ a n + 1 a n ∣ = ∣ x n + 1 / ( n + 1 ) ! x n / n ! ∣ = ∣ x ∣ n + 1 . \left|\frac{a_{n+1}}{a_n}\right|
=
\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right|
=
\frac{|x|}{n+1}. a n a n + 1 = x n / n ! x n + 1 / ( n + 1 )! = n + 1 ∣ x ∣ . Take the limit:
lim n → ∞ ∣ x ∣ n + 1 = 0. \lim_{n\to\infty}\frac{|x|}{n+1}=0. n → ∞ lim n + 1 ∣ x ∣ = 0. Since this limit is 0<1 for every real x x x x x x
R = ∞ , interval of convergence ( − ∞ , ∞ ) . R=\infty,
\quad
\text{interval of convergence } (-\infty,\infty). R = ∞ , interval of convergence ( − ∞ , ∞ ) . Exercise 2
Find all x x x
∑ n = 0 ∞ n ! x n . \sum_{n=0}^{\infty} n! x^n. n = 0 ∑ ∞ n ! x n . Show solution ↓ Hide solution ↑ Solution
Let a n = n ! x n a_n=n!x^n a n = n ! x n
Apply the Ratio Test:
∣ a n + 1 a n ∣ = ∣ ( n + 1 ) ! x n + 1 n ! x n ∣ = ( n + 1 ) ∣ x ∣ . \left|\frac{a_{n+1}}{a_n}\right|
=
\left|\frac{(n+1)!x^{n+1}}{n!x^n}\right|
=
(n+1)|x|. a n a n + 1 = n ! x n ( n + 1 )! x n + 1 = ( n + 1 ) ∣ x ∣. Take the limit:
lim n → ∞ ( n + 1 ) ∣ x ∣ . \lim_{n\to\infty}(n+1)|x|. n → ∞ lim ( n + 1 ) ∣ x ∣. If x ≠ 0 x\neq0 x = 0 n + 1 → ∞ n+1\to\infty n + 1 → ∞ x ≠ 0 x\neq0 x = 0
If x = 0 x=0 x = 0 n = 0 n=0 n = 0
R = 0 , converges only at x = 0. R=0,
\quad
\text{converges only at } x=0. R = 0 , converges only at x = 0. Exercise 3
Find the interval of convergence:
∑ n = 1 ∞ ( − 1 ) n x n n ( n + 1 ) . \sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n(n+1)}. n = 1 ∑ ∞ n ( n + 1 ) ( − 1 ) n x n . Show solution ↓ Hide solution ↑ Solution
Let a n = ( − 1 ) n x n n ( n + 1 ) a_n=\frac{(-1)^n x^n}{n(n+1)} a n = n ( n + 1 ) ( − 1 ) n x n
Apply the Ratio Test:
∣ a n + 1 a n ∣ = ∣ ( − 1 ) n + 1 x n + 1 / [ ( n + 1 ) ( n + 2 ) ] ( − 1 ) n x n / [ n ( n + 1 ) ] ∣ = ∣ x ∣ n n + 2 . \left|\frac{a_{n+1}}{a_n}\right|
=
\left|
\frac{(-1)^{n+1}x^{n+1}/[(n+1)(n+2)]}
{(-1)^n x^n/[n(n+1)]}
\right|
=
|x|\frac{n}{n+2}. a n a n + 1 = ( − 1 ) n x n / [ n ( n + 1 )] ( − 1 ) n + 1 x n + 1 / [( n + 1 ) ( n + 2 )] = ∣ x ∣ n + 2 n . Now take the limit:
lim n → ∞ ∣ x ∣ n n + 2 = ∣ x ∣ . \lim_{n\to\infty}|x|\frac{n}{n+2}
=
|x|. n → ∞ lim ∣ x ∣ n + 2 n = ∣ x ∣. The Ratio Test gives convergence when
|x|<1. Thus the radius of convergence is R = 1 R=1 R = 1
Now test endpoints.
At x = 1 x=1 x = 1
∑ n = 1 ∞ ( − 1 ) n n ( n + 1 ) . \sum_{n=1}^{\infty} \frac{(-1)^n}{n(n+1)}. n = 1 ∑ ∞ n ( n + 1 ) ( − 1 ) n . Consider b n = 1 n ( n + 1 ) b_n=\frac{1}{n(n+1)} b n = n ( n + 1 ) 1
b_n>0, \quad b_{n+1}<b_n, \quad \lim_{n\to\infty}b_n=0. Thus the series satisfies the conditions of the Alternating Series Test , so it converges (conditionally).
At x = − 1 x=-1 x = − 1
∑ n = 1 ∞ 1 n ( n + 1 ) . \sum_{n=1}^{\infty} \frac{1}{n(n+1)}. n = 1 ∑ ∞ n ( n + 1 ) 1 . Notice
1 n ( n + 1 ) = 1 n − 1 n + 1 . \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}. n ( n + 1 ) 1 = n 1 − n + 1 1 . This is a telescoping series. The partial sums become
( 1 − 1 2 ) + ( 1 2 − 1 3 ) + ( 1 3 − 1 4 ) + ⋯ \left(1-\frac12\right)
+
\left(\frac12-\frac13\right)
+
\left(\frac13-\frac14\right)
+\cdots ( 1 − 2 1 ) + ( 2 1 − 3 1 ) + ( 3 1 − 4 1 ) + ⋯ Everything cancels except the first and last terms, so the series converges.
Therefore,
Interval of convergence [ − 1 , 1 ] . \boxed{\text{Interval of convergence } [-1,1]}. Interval of convergence [ − 1 , 1 ] . Exercise 4
Find the interval of convergence:
∑ n = 1 ∞ ( x − 2 ) n n 3 n . \sum_{n=1}^{\infty} \frac{(x-2)^n}{n\,3^n}. n = 1 ∑ ∞ n 3 n ( x − 2 ) n . Show solution ↓ Hide solution ↑ Solution
Let a n = ( x − 2 ) n n 3 n a_n=\frac{(x-2)^n}{n3^n} a n = n 3 n ( x − 2 ) n
Apply the Ratio Test:
∣ a n + 1 a n ∣ = ∣ ( x − 2 ) n + 1 / [ ( n + 1 ) 3 n + 1 ] ( x − 2 ) n / [ n 3 n ] ∣ = ∣ x − 2 ∣ 3 n n + 1 . \left|\frac{a_{n+1}}{a_n}\right|
=
\left|
\frac{(x-2)^{n+1}/[(n+1)3^{n+1}]}
{(x-2)^n/[n3^n]}
\right|
=
\frac{|x-2|}{3}\frac{n}{n+1}. a n a n + 1 = ( x − 2 ) n / [ n 3 n ] ( x − 2 ) n + 1 / [( n + 1 ) 3 n + 1 ] = 3 ∣ x − 2∣ n + 1 n . Take the limit:
lim n → ∞ ∣ x − 2 ∣ 3 n n + 1 = ∣ x − 2 ∣ 3 . \lim_{n\to\infty}
\frac{|x-2|}{3}\frac{n}{n+1}
=
\frac{|x-2|}{3}. n → ∞ lim 3 ∣ x − 2∣ n + 1 n = 3 ∣ x − 2∣ . Thus convergence occurs when
\frac{|x-2|}{3}<1
\quad\Rightarrow\quad
|x-2|<3. This gives
-1<x<5. Now test endpoints.
At x = 5 x=5 x = 5
∑ n = 1 ∞ 3 n n 3 n = ∑ n = 1 ∞ 1 n . \sum_{n=1}^{\infty}\frac{3^n}{n3^n}
=
\sum_{n=1}^{\infty}\frac1n. n = 1 ∑ ∞ n 3 n 3 n = n = 1 ∑ ∞ n 1 . This is the harmonic series , which diverges.
At x = − 1 x=-1 x = − 1
∑ n = 1 ∞ ( − 3 ) n n 3 n = ∑ n = 1 ∞ ( − 1 ) n n . \sum_{n=1}^{\infty}\frac{(-3)^n}{n3^n}
=
\sum_{n=1}^{\infty}\frac{(-1)^n}{n}. n = 1 ∑ ∞ n 3 n ( − 3 ) n = n = 1 ∑ ∞ n ( − 1 ) n . This is the alternating harmonic series .
The terms decrease to zero and alternate in sign, so by the Alternating Series Test it converges (conditionally).
Therefore,
Interval of convergence [ − 1 , 5 ) . \boxed{\text{Interval of convergence } [-1,5)}. Interval of convergence [ − 1 , 5 ) .