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Exercises: Power Series

Theory

Find the radius and interval of convergence.

Exercises

Exercise 1
∑n=0∞xnn+2\sum_{n=0}^{\infty} \frac{x^{n}}{n+2}
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Solution

Let an=xnn+2a_n=\frac{x^n}{n+2}.

Ratio Test:

∣an+1an∣=∣xn+1n+3⋅n+2xn∣=∣x∣n+2n+3.\left|\frac{a_{n+1}}{a_n}\right| = \left|\frac{x^{n+1}}{n+3}\cdot\frac{n+2}{x^n}\right| = |x|\frac{n+2}{n+3}.lim⁡n→∞∣x∣n+2n+3=∣x∣.\lim_{n\to\infty}|x|\frac{n+2}{n+3}=|x|.

Converges when |x|<1, so R=1R=1.

Endpoints:

At x=1x=1:

∑1n+2\sum \frac{1}{n+2}

behaves like ∑1n\sum \frac1n, a harmonic series, so diverges by comparison.

At x=−1x=-1:

∑(−1)nn+2\sum \frac{(-1)^n}{n+2}

is alternating, terms decrease to 00, so converges by Alternating Series Test.

R=1,[−1,1).R=1,\quad [-1,1).
Exercise 2
∑n=0∞nxn\sum_{n=0}^{\infty} nx^n
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Solution

Ratio Test:

∣(n+1)xn+1nxn∣=∣x∣n+1n→∣x∣.\left|\frac{(n+1)x^{n+1}}{nx^n}\right| = |x|\frac{n+1}{n} \to |x|.

Thus |x|<1, so R=1R=1.

Endpoints:

x=1x=1: ∑n\sum n diverges.

x=−1x=-1: ∑n(−1)n\sum n(-1)^n does not approach 00, so diverges by Divergence Test.

R=1,(−1,1).R=1,\quad (-1,1).
Exercise 3
∑n=0∞xnn!\sum_{n=0}^{\infty} \frac{x^n}{n!}
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Solution

Ratio Test:

∣xn+1∣→0.\left|\frac{x}{n+1}\right|\to 0.

Always less than 1.

R=∞,(−∞,∞).R=\infty,\quad (-\infty,\infty).
Exercise 4
∑n=1∞(−1)nxnn2n\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n2^n}
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Solution

Ratio Test:

∣x2∣.\left|\frac{x}{2}\right|.

Thus |x|<2, so R=2R=2.

Endpoints:

x=2x=2:

∑(−1)nn\sum \frac{(-1)^n}{n}

alternating harmonic series → converges.

x=−2x=-2:

∑1n\sum \frac{1}{n}

harmonic series → diverges.

R=2,(−2,2].R=2,\quad (-2,2].
Exercise 5
∑n=0∞3nxn(n+1)2\sum_{n=0}^{\infty} \frac{3^n x^n}{(n+1)^2}
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Solution

Ratio Test:

∣3x∣.|3x|.

So |3x|<1 \Rightarrow |x|<\frac13, hence R=13R=\frac13.

Endpoints:

x=13x=\frac13:

∑1(n+1)2\sum \frac{1}{(n+1)^2}

is a pp-series with p=2>1, converges.

x=−13x=-\frac13:

∑(−1)n(n+1)2\sum \frac{(-1)^n}{(n+1)^2}

absolutely convergent since ∑1(n+1)2\sum \frac1{(n+1)^2} converges.

R=13,[−13,13].R=\frac13,\quad \left[-\frac13,\frac13\right].
Exercise 6
∑n=2∞xnln⁡n\sum_{n=2}^{\infty} \frac{x^n}{\ln n}
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Solution

Root Test:

1ln⁡nn→1.\sqrt[n]{\frac1{\ln n}} \to 1.

So |x|<1, R=1R=1.

Endpoints:

x=1x=1:

∑1ln⁡n\sum \frac1{\ln n}

diverges by comparison with 1n\frac1n (since \ln n < n for large nn).

x=−1x=-1:

∑(−1)nln⁡n\sum \frac{(-1)^n}{\ln n}

terms decrease to 0 and alternate → converges by Alternating Series Test.

R=1,[−1,1).R=1,\quad [-1,1).
Exercise 7
∑n=0∞44n(2x−1)n\sum_{n=0}^{\infty} \frac{4}{4^n}(2x-1)^n
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Solution

Geometric form:

\left|\frac{2x-1}{4}\right|<1.

So |2x-1|<4.

-\frac32 < x < \frac52.R=2.R=2.
Exercise 8
∑n=1∞(−1)n(x−1)nn\sum_{n=1}^{\infty} \frac{(-1)^n (x-1)^n}{\sqrt{n}}
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Solution

Root Test:

1nn→1.\sqrt[n]{\frac1{\sqrt n}}\to 1.

Thus |x-1|<1, R=1R=1.

Endpoints:

x=0x=0 and x=2x=2 give

∑(−1)nn.\sum \frac{(-1)^n}{\sqrt n}.

Alternating, decreasing, limit 0 → converges.

Not absolutely convergent (since ∑1/n\sum 1/\sqrt n diverges).

R=1,[0,2].R=1,\quad [0,2].
Exercise 9
∑n=1∞(x−2)nnn\sum_{n=1}^{\infty} \frac{(x-2)^n}{n^n}
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Solution

Let

an=(x−2)nnn.a_n = \frac{(x-2)^n}{n^n}.

Apply the Root Test:

lim⁡n→∞∣an∣n=lim⁡n→∞∣x−2∣n.\lim_{n\to\infty} \sqrt[n]{|a_n|} = \lim_{n\to\infty} \frac{|x-2|}{n}.

Since ∣x−2∣n→0\frac{|x-2|}{n} \to 0 for every real xx, the limit is 0<1.

Therefore, the series converges absolutely for all xx.

R=∞\boxed{R=\infty}Interval of convergence (−∞,∞)\boxed{\text{Interval of convergence } (-\infty,\infty)}
Exercise 10
∑n=0∞2n(x−3)nn+3\sum_{n=0}^{\infty} \frac{2^n (x-3)^n}{n+3}
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Solution

Let

an=2n(x−3)nn+3.a_n=\frac{2^n (x-3)^n}{n+3}.

Apply the Ratio Test:

∣an+1an∣=2n+1∣x−3∣n+1n+4⋅n+32n∣x−3∣n=2∣x−3∣n+3n+4.\left| \frac{a_{n+1}}{a_n} \right| = \frac{2^{n+1}|x-3|^{n+1}}{n+4} \cdot \frac{n+3}{2^n |x-3|^n} = 2|x-3|\frac{n+3}{n+4}.

Take the limit:

lim⁡n→∞2∣x−3∣n+3n+4=2∣x−3∣.\lim_{n\to\infty} 2|x-3|\frac{n+3}{n+4} = 2|x-3|.

For convergence:

2|x-3|<1 \quad \Rightarrow \quad |x-3|<\frac12.

Thus

R=12.R=\frac12.

Now test endpoints.

At x=72x=\frac72:

∑1n+3\sum \frac{1}{n+3}

behaves like the harmonic series ∑1n\sum \frac1n, which diverges.

At x=52x=\frac52:

∑(−1)nn+3\sum \frac{(-1)^n}{n+3}

is an alternating series with decreasing terms approaching 00, so it converges by the Alternating Series Test.

R=12\boxed{R=\frac12}[52,72)\boxed{\left[\frac52,\frac72\right)}
Exercise 11
∑n=0∞n(n2+1)4n(x−10)n\sum_{n=0}^{\infty} \frac{n}{(n^2+1)4^n}(x-10)^n
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Solution

Apply the Root Test:

∣n(n2+1)4n(x−10)n∣n=∣x−10∣4nn2+1n.\sqrt[n]{\left|\frac{n}{(n^2+1)4^n}(x-10)^n\right|} = \frac{|x-10|}{4} \sqrt[n]{\frac{n}{n^2+1}}.

Since

nn2+1n→1,\sqrt[n]{\frac{n}{n^2+1}} \to 1,

the limit becomes

∣x−10∣4.\frac{|x-10|}{4}.

Thus convergence requires

\frac{|x-10|}{4}<1 \quad \Rightarrow \quad |x-10|<4.

So

R=4.R=4.

Endpoints:

At x=14x=14 or x=6x=6 the series behaves like

∑nn2+1,\sum \frac{n}{n^2+1},

which behaves like ∑1n\sum \frac1n and diverges by comparison with the harmonic series.

R=4\boxed{R=4}(6,14)\boxed{(6,14)}
Exercise 12
∑n=1∞(n2)n(x+6)n\sum_{n=1}^{\infty} \left(\frac{n}{2}\right)^n (x+6)^n
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Solution

Apply the Root Test:

(n2)n∣x+6∣nn=n2∣x+6∣.\sqrt[n]{\left(\frac{n}{2}\right)^n |x+6|^n} = \frac{n}{2}|x+6|.

As n→∞n\to\infty, this limit is infinite unless ∣x+6∣=0|x+6|=0.

Thus convergence occurs only when x=−6x=-6.

R=0\boxed{R=0}Converges only at x=−6\boxed{\text{Converges only at } x=-6}
Exercise 13
∑n=1∞(2x−1)nn3\sum_{n=1}^{\infty} \frac{(2x-1)^n}{n^3}
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Solution

Ratio Test:

∣an+1an∣=∣2x−1∣n3(n+1)3.\left|\frac{a_{n+1}}{a_n}\right| = |2x-1| \frac{n^3}{(n+1)^3}.

Taking limit:

∣2x−1∣.|2x-1|.

Thus

|2x-1|<1 \quad \Rightarrow \quad 0<x<1.

Radius:

R=12.R=\frac12.

Endpoints:

At x=0x=0 or x=1x=1:

∑1n3\sum \frac{1}{n^3}

is a pp-series with p=3>1, so it converges.

R=12\boxed{R=\frac12}[0,1]\boxed{[0,1]}
Exercise 14
∑n=0∞nn+1(x−e)n\sum_{n=0}^{\infty} \frac{n}{\sqrt{n+1}}(x-e)^n
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Solution

Root Test:

nn+1n∣x−e∣→∣x−e∣.\sqrt[n]{\frac{n}{\sqrt{n+1}}}|x-e| \to |x-e|.

Thus

|x-e|<1.R=1.R=1.

Endpoints give terms behaving like ∑n\sum \sqrt n, which diverges by comparison with ∑1\sum 1.

R=1\boxed{R=1}(e−1,e+1)\boxed{(e-1,e+1)}
Exercise 15
∑n=1∞2⋅4⋅6⋯(2n)1⋅3⋅5⋯(2n−1)xn\sum_{n=1}^{\infty} \frac{2\cdot4\cdot6\cdots(2n)}{1\cdot3\cdot5\cdots(2n-1)}x^n
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Solution

Ratio Test gives limit

∣x∣.|x|.

Thus

|x|<1.R=1.R=1.

Endpoints fail the Test for Divergence since terms do not approach 00.

(−1,1)\boxed{(-1,1)}
Exercise 16
∑n=1∞nxn1⋅3⋅5⋯(2n−1)\sum_{n=1}^{\infty} \frac{nx^n}{1\cdot3\cdot5\cdots(2n-1)}
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Solution

Root Test gives limit 00 for every xx because the denominator grows super-exponentially.

R=∞\boxed{R=\infty}(−∞,∞)\boxed{(-\infty,\infty)}
Exercise 17
∑n=2∞(−1)n(2x+3)nnln⁡n\sum_{n=2}^{\infty} \frac{(-1)^n (2x+3)^n}{n\ln n}
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Solution

Ratio Test gives

|2x+3|<1.

Thus

-2<x<-1.R=12.R=\frac12.

Endpoints:

At x=−1x=-1: alternating series with decreasing terms →0\to 0, converges.

At x=−2x=-2: series becomes ∑1nln⁡n\sum \frac1{n\ln n}, which diverges by integral test.

(−2,−1]\boxed{(-2,-1]}
Exercise 18
∑n=0∞xn(ln⁡n)n\sum_{n=0}^{\infty} \frac{x^n}{(\ln n)^n}
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Solution

Root Test:

∣x∣n(ln⁡n)nn=∣x∣ln⁡n→0.\sqrt[n]{\frac{|x|^n}{(\ln n)^n}} = \frac{|x|}{\ln n} \to 0.

Thus converges for all xx.

R=∞\boxed{R=\infty}(−∞,∞)\boxed{(-\infty,\infty)}
Exercise 19
∑n=0∞n!10n(x−π)n\sum_{n=0}^{\infty} \frac{n!}{10^n}(x-\pi)^n
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Solution

Ratio Test:

∣an+1an∣=(n+1)!10n+110nn!∣x−π∣=n+110∣x−π∣.\left|\frac{a_{n+1}}{a_n}\right| = \frac{(n+1)!}{10^{n+1}} \frac{10^n}{n!} |x-\pi| = \frac{n+1}{10}|x-\pi|.

Limit is infinite unless ∣x−π∣=0|x-\pi|=0.

R=0\boxed{R=0}Converges only at x=π\boxed{\text{Converges only at } x=\pi}
Exercise 20
∑n=1∞(−1)nx2n−1(2n−1)!\sum_{n=1}^{\infty} \frac{(-1)^n x^{2n-1}}{(2n-1)!}
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Solution

Ratio Test gives limit

lim⁡n→∞∣x∣2(2n)(2n+1)=0\lim_{n\to\infty} \frac{|x|^2}{(2n)(2n+1)}=0

for every xx.

Thus converges for all xx.

R=∞\boxed{R=\infty}(−∞,∞)\boxed{(-\infty,\infty)}