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Week 6 Discussion Notes

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Power Series

Power series are a special type of infinite series:

Definition

Let F(x)=∑n=0∞an(x−c)n\displaystyle F\p{x} = \sum_{n=0}^\infty a_n\p{x - c}^n. FF is called a power series and we call cc the center of FF.

Power series are great because they have a bunch of nice properties (during discussion, I only talked about the first property):

Proposition (properties of power series)

Let F(x)=∑n=0∞an(x−c)n\displaystyle F\p{x} = \sum_{n=0}^\infty a_n\p{x - c}^n be a power series.

  1. There exists a number R∈[0,∞]R \in \br{0, \infty} called the radius of convergence of FF such that if ∣x−c∣<R\abs{x - c} < R, then FF converges absolutely and if ∣x−c∣>R\abs{x - c} > R, then FF diverges. If R=0R = 0, then the series only converges at its center and if R=∞R = \infty, then FF converges absolutely everywhere.

  2. The coefficients can be expressed in terms of the derivatives of FF:

    an=F(n)(c)n!a_n = \frac{F^{\p{n}}\p{c}}{n!}

  3. You can differentiate term-by-term, i.e., you can pretend the sum isn't there and take the derivative like normal:

    F′(x)=∑n=1∞nan(x−c)n−1F'\p{x} = \sum_{n=1}^\infty na_n\p{x - c}^{n-1}

    F′F' is also a power series and it has the same radius of convergence as FF (but not necessarily the same interval of convergence).

  4. You can also integrate term-by-term:

    ∫F(x) dx=C+∑n=0∞ann+1(x−c)n+1\int F\p{x} \,\diff{x} = C + \sum_{n=0}^\infty \frac{a_n}{n+1} \p{x - c}^{n+1}

    Like F′F', the antiderivative of FF is a power series with the same radius of convergence as FF (but not necessarily the same interval of convergence).
Example 1.

We've already seen an example of a power series: if we replace rr with xx in the geometric series, then we get

∑n=0∞rn=11−r⇝∑n=0∞xn=11−x,\sum_{n=0}^\infty r^n = \frac{1}{1 - r} \rightsquigarrow \sum_{n=0}^\infty x^n = \frac{1}{1 - x},

and this converges if and only if ∣x∣<1\abs{x} < 1. Thus, the geometric series above is an example of a power series centered at 00 and with interval of convergence (−1,1)\p{-1, 1}.

A common type of problem involving power series is finding its interval of convergence:

Example 2.

Find the interval of convergence of ∑n=1∞3nxnn3\displaystyle\sum_{n=1}^\infty \frac{3^n x^n}{n^3}.
Solution.

For these types of questions, there's a step-by-step way to solve them:

  1. Use the ratio or root test to find the radius of convergence.
  2. Check whether the series converges at the endpoints.

For this series, we'll use the ratio test:

lim⁡n→∞∣an+1an∣=lim⁡n→∞3n+1∣x∣n+1(n+1)3n33n∣x∣n=lim⁡n→∞3∣x∣(nn+1)3=3∣x∣.\begin{aligned} \lim_{n\to\infty} \abs{\frac{a_{n+1}}{a_n}} &= \lim_{n\to\infty} \frac{3^{n+1} \abs{x}^{n+1}}{\p{n+1}^3} \frac{n^3}{3^n \abs{x}^n} \\ &= \lim_{n\to\infty} 3 \abs{x} \p{\frac{n}{n+1}}^3 \\ &= 3\abs{x}. \end{aligned}

The ratio test tells us that the series will converge absolutely if

3∣x∣<1  ⟺  ∣x∣<13  ⟺  x∈(−13,13).3\abs{x} < 1 \iff \abs{x} < \frac{1}{3} \iff x \in \p{-\frac{1}{3}, \frac{1}{3}}.

This tells us that the radius of convergence is 13\frac{1}{3}. Now we need to check the endpoints:

x=−13x = -\frac{1}{3}:

We get

∑n=1∞3n(−13)nn3=∑n=1∞(−1)nn3,\sum_{n=1}^\infty \frac{3^n \p{-\frac{1}{3}}^n}{n^3} = \sum_{n=1}^\infty \frac{\p{-1}^n}{n^3},

which converges, for example, by the alternating series test. (Alternatively, you can say that it converges absolutely by the pp-test.)

x=13x = \frac{1}{3}:

We get a similar series in this case:

∑n=1∞3n(13)nn3=∑n=1∞1n3,\sum_{n=1}^\infty \frac{3^n \p{\frac{1}{3}}^n}{n^3} = \sum_{n=1}^\infty \frac{1}{n^3},

and this converges by the pp-test.

Thus, the interval of convergence of this power series is

[−13,13].\boxed{\br{-\frac{1}{3}, \frac{1}{3}}}.

Convergence Tests

Example 3.

Determine whether ∑n=0∞nnn4+ln⁡n\displaystyle \sum_{n=0}^\infty \frac{n\sqrt{n}}{n^4 + \ln{n}} converges.
Solution.

Heuristically, ln⁡n\ln{n} grows very, very slow, so if nn is very large,

nnn4+ln⁡n≈nnn4=1n5/2.\frac{n\sqrt{n}}{n^4 + \ln{n}} \approx \frac{n\sqrt{n}}{n^4} = \frac{1}{n^{5/2}}.

So, we'll try to apply the limit test with bn=1n5/2b_n = \frac{1}{n^{5/2}}, which is summable by the pp-test.

lim⁡n→∞nnn4+ln⁡n1n5/2=lim⁡n→∞n4n4+ln⁡n⋅1n41n4=lim⁡n→∞11+ln⁡nn4=1,\begin{aligned} \lim_{n\to\infty} \frac{\frac{n\sqrt{n}}{n^4 + \ln{n}}}{\frac{1}{n^{5/2}}} &= \lim_{n\to\infty} \frac{n^4}{n^4 + \ln{n}} \cdot \frac{\frac{1}{n^4}}{\frac{1}{n^4}} \\ &= \lim_{n\to\infty} \frac{1}{1 + \frac{\ln{n}}{n^4}} \\ &= 1, \end{aligned}

so the limit test tells us that because ∑n=1∞1n5/2\sum_{n=1}^\infty \frac{1}{n^{5/2}} converges, ∑n=1∞nnn4+ln⁡n\sum_{n=1}^\infty \frac{n\sqrt{n}}{n^4 + \ln{n}} also converges.
Example 4.

Determine whether ∑n=0∞cos⁡nπn\displaystyle \sum_{n=0}^\infty \frac{\cos{n\pi}}{\sqrt{n}} converges.
Solution.

The trick for this is to notice that cos⁡nπ=(−1)n\cos{n\pi} = \p{-1}^n, so the series is

∑n=0∞cos⁡nπn=∑n=0∞(−1)nn,\sum_{n=0}^\infty \frac{\cos{n\pi}}{\sqrt{n}} = \sum_{n=0}^\infty \frac{\p{-1}^n}{\sqrt{n}},

and this converges by the alternating series test.
Example 5.

Determine whether ∑n=0∞nsin⁡(1n2)\displaystyle \sum_{n=0}^\infty n\sin\p{\frac{1}{n^2}} converges.
Solution.

Recall from last week the fact that lim⁡x→0sin⁡xx=1\lim_{x\to0} \frac{\sin{x}}{x} = 1. Heuristically, this means that if x≈0x \approx 0, then

sin⁡xx≈1  ⟹  sin⁡x≈x.\frac{\sin{x}}{x} \approx 1 \implies \sin{x} \approx x.

Specializing to our problem, if nn is large enough, then 1n2≈0\frac{1}{n^2} \approx 0, so if we plug in x=1n2x = \frac{1}{n^2}, we get

sin⁡(1n2)≈1n2  ⟹  nsin⁡(1n2)≈1n.\sin\p{\frac{1}{n^2}} \approx \frac{1}{n^2} \implies n\sin\p{\frac{1}{n^2}} \approx \frac{1}{n}.

This tells us to try bn=1nb_n = \frac{1}{n} with limit comparison:

lim⁡n→∞nsin⁡(1n2)1n=lim⁡n→∞sin⁡(1n2)1n2=1.\lim_{n\to\infty} \frac{n\sin\p{\frac{1}{n^2}}}{\frac{1}{n}} = \lim_{n\to\infty} \frac{\sin\p{\frac{1}{n^2}}}{\frac{1}{n^2}} = 1.

But ∑n=1∞1n\sum_{n=1}^\infty \frac{1}{n} diverges by the pp-test, so by the limit comparison test, ∑n=1∞nsin⁡(1n2)\sum_{n=1}^\infty n\sin\p{\frac{1}{n^2}} also diverges.
Example 6.

Determine whether ∑n=0∞(1−cos⁡(1n))\displaystyle \sum_{n=0}^\infty \p{1 - \cos\p{\frac{1}{n}}} converges.
Solution.

Like the previous example, we'll use a "well-known" limit to do this problem:

lim⁡x→01−cos⁡xx2=12.\lim_{x\to0} \frac{1 - \cos{x}}{x^2} = \frac{1}{2}.

You can prove this as follows:

lim⁡x→01−cos⁡xx2⋅1+cos⁡x1+cos⁡x=lim⁡x→0sin⁡2xx211+cos⁡x=lim⁡x→0(sin⁡xx)211+cos⁡x=12.\begin{aligned} \lim_{x\to0} \frac{1 - \cos{x}}{x^2} \cdot \frac{1 + \cos{x}}{1 + \cos{x}} &= \lim_{x\to0} \frac{\sin^2{x}}{x^2} \frac{1}{1 + \cos{x}} \\ &= \lim_{x\to0} \p{\frac{\sin{x}}{x}}^2 \frac{1}{1 + \cos{x}} \\ &= \frac{1}{2}. \end{aligned}

So, we get the following heuristic: if x≈0x \approx 0, then

1−cos⁡xx2≈12  ⟹  1−cos⁡x≈x22.\frac{1 - \cos{x}}{x^2} \approx \frac{1}{2} \implies 1 - \cos{x} \approx \frac{x^2}{2}.

If we plug in x=1nx = \frac{1}{n}, then

1−cos⁡(1n)≈12n2.1 - \cos\p{\frac{1}{n}} \approx \frac{1}{2n^2}.

From here, you can use bn=1n2b_n = \frac{1}{n^2} (or bn=12n2b_n = \frac{1}{2n^2}; both will work) in the limit comparison test. Then

lim⁡n→∞1−cos⁡(1n)1n2=lim⁡n→∞1−cos⁡(1n)(1n)2=12.\lim_{n\to\infty} \frac{1 - \cos\p{\frac{1}{n}}}{\frac{1}{n^2}} = \lim_{n\to\infty} \frac{1 - \cos\p{\frac{1}{n}}}{\p{\frac{1}{n}}^2} = \frac{1}{2}.

The pp-test tells us that ∑n=1∞1n2\sum_{n=1}^\infty \frac{1}{n^2} converges, so by limit comparison, ∑n=1∞(1−cos⁡(1n))\sum_{n=1}^\infty \p{1 - \cos\p{\frac{1}{n}}} also converges.
Example 7.

True or false:

If ∑nan\sum_n a_n converges absolutely, then ∑nan2\sum_n a_n^2 converges.
Solution.

This is true. Since ∑n∣an∣\sum_n \abs{a_n} converges, we know that

lim⁡n→∞∣an∣=0.\lim_{n\to\infty} \abs{a_n} = 0.

This means that eventually, ∣an∣≤1\abs{a_n} \leq 1. Mathematically, this means there exists N>0N > 0 such that if n≥Nn \geq N, then ∣an∣≤1\abs{a_n} \leq 1. So, if n≥Nn \geq N, then

an2=∣an∣2=∣an∣⋅∣an∣≤1⋅∣an∣=∣an∣.\begin{aligned} a_n^2 &= \abs{a_n}^2 \\ &= \abs{a_n} \cdot \abs{a_n} \\ &\leq 1 \cdot \abs{a_n} \\ &= \abs{a_n}. \end{aligned}

Since ∑n=N∞∣an∣\sum_{n=N}^\infty \abs{a_n} converges, direct comparison tells us that ∑n=N∞an2\sum_{n=N}^\infty a_n^2 also converges, so

∑n=1∞an2=∑n=1N−1an2+∑n=N∞an2,\sum_{n=1}^\infty a_n^2 = \sum_{n=1}^{N-1} a_n^2 + \sum_{n=N}^\infty a_n^2,

i.e., this is the sum of two convergent series, so the entire series converges.