Working with Power Series
Section 6.2: partial fractions, products, differentiating & integrating; how to work the HW 6.2 problems
Last time we learned that
\[ \frac{1}{1-u} = \sum_{n=0}^\infty u^n, \qquad |u| < 1, \]
and that substituting for \(u\) gives series for things like \(\frac{1}{1+x^2}\) or \(\frac{1}{1-4x}\). This section is about getting many more series out of that one formula, using algebra and calculus.
1. Partial fractions first, then expand (#1)
A rational function with a factored denominator is not geometric as it stands. Split it with partial fractions, then expand each piece separately.
Example. \(\displaystyle f(x) = \frac{3}{(x-2)(x+1)}\)
Partial fractions: \(\dfrac{3}{(x-2)(x+1)} = \dfrac{A}{x-2} + \dfrac{B}{x+1}\). Cover-up method: \(A = \frac{3}{2+1} = 1\) and \(B = \frac{3}{-1-2} = -1\).
Expand each piece. The constant in the denominator must be turned into a \(1\):
\[ \frac{1}{x-2} = \frac{-1}{2 - x} = -\frac12\cdot\frac{1}{1 - \frac x2} = -\sum_{n=0}^\infty \frac{x^n}{2^{n+1}}, \qquad |x| < 2 \]
\[ \frac{-1}{x+1} = -\frac{1}{1-(-x)} = -\sum_{n=0}^\infty (-1)^n x^n = \sum_{n=0}^\infty (-1)^{n+1}x^n, \qquad |x| < 1 \]
Combine into one sum (both start at \(n = 0\) with \(x^n\)):
\[ f(x) = \sum_{n=0}^\infty \left[(-1)^{n+1} - \frac{1}{2^{n+1}}\right]x^n \]
Check: \(f(0) = \frac{3}{(-2)(1)} = -\frac32\), and the \(n = 0\) coefficient is \(-1 - \frac12 = -\frac32\). ✓
The series is valid where both pieces are valid: \(|x| < 1\).
Write the answer as \(c\left[(\ldots)^{n+1} - (\ldots)^{n+1}\right]x^n\) with fractions like (1/2)^(n+1) and (-1)^(n+1). Decimals like 0.5 are accepted too.
2. Going backwards: series → rational function (#2, 3)
If a series is geometric, it equals \(\dfrac{\text{first term}}{1 - \text{ratio}}\). Watch where the index starts.
Example. \(\displaystyle\sum_{n=1}^\infty \frac{1}{x^{3n}}\). The first term (\(n = 1\)) is \(\frac{1}{x^3}\) and the ratio is \(\frac{1}{x^3}\):
\[ \sum_{n=1}^\infty \frac{1}{x^{3n}} = \frac{1/x^3}{1 - 1/x^3} = \frac{1}{x^3 - 1} \qquad \left(\text{valid for } |x| > 1\right) \]
(Multiply top and bottom by \(x^3\) in the last step.)
Periodic sign patterns: group the terms. Consider
\[ 1 + 2x - x^2 + x^3 + 2x^4 - x^5 + x^6 + 2x^7 - x^8 + \cdots \]
The coefficients repeat \(1, 2, -1\) every three terms. Group them in threes and factor:
\[ (1 + 2x - x^2) + x^3(1 + 2x - x^2) + x^6(1 + 2x - x^2) + \cdots = (1 + 2x - x^2)\sum_{k=0}^\infty x^{3k} \]
\[ = \frac{1 + 2x - x^2}{1 - x^3}, \qquad |x| < 1 \]
Write one block of \(m\) terms, \(p(x)\), and factor it out. What’s left is \(\sum (x^m)^k = \frac{1}{1 - x^m}\), so the sum is \(\dfrac{p(x)}{1-x^m}\).
If the series starts at \(x^1\) instead of \(x^0\), the first block simply starts with \(x\).
3. Multiplying two series (#4)
To multiply \(\sum a_n x^n\) by \(\sum b_n x^n\), collect every pair of terms whose powers add up to \(n\):
\[ \left(\sum_{n=0}^\infty a_n x^n\right)\left(\sum_{n=0}^\infty b_n x^n\right) = \sum_{n=0}^\infty c_n x^n, \qquad c_n = a_0 b_n + a_1 b_{n-1} + \cdots + a_n b_0 = \sum_{k=0}^n a_k b_{n-k} \]
This works just like multiplying polynomials (FOIL, then group like powers), only with infinitely many terms. It’s called the Cauchy product.
Example. \(f(x) = \sum 2x^n\) and \(g(x) = \sum x^n\). Every \(a_k = 2\) and every \(b_{n-k} = 1\), so
\[ c_n = \sum_{k=0}^n 2\cdot1 = 2(n+1), \qquad f(x)g(x) = \sum_{n=0}^\infty 2(n+1)x^n \]
Sanity check: \(f g = \frac{2}{1-x}\cdot\frac{1}{1-x} = \frac{2}{(1-x)^2}\). We’ll see in §4 that this equals \(\sum 2(n+1)x^n\). ✓
Example. \(\frac{1}{1-x}\cdot\frac{1}{1+x}\), so \(a_k = 1\) and \(b_k = (-1)^k\):
\[ c_n = \sum_{k=0}^n (-1)^{n-k} = \begin{cases}1 & n \text{ even}\\ 0 & n\text{ odd}\end{cases} \quad\Longrightarrow\quad \frac{1}{1-x^2} = 1 + x^2 + x^4 + \cdots \]
which matches the substitution \(u = x^2\). ✓
\(\displaystyle\sum_{k=0}^n 1 = n+1\), \(\qquad\displaystyle\sum_{k=0}^n k = \frac{n(n+1)}{2}\)
4. Differentiating term by term (#5, 10)
If \(f(x) = \sum c_n x^n\) has radius \(R > 0\), then for \(|x| < R\)
\[ f'(x) = \sum_{n=1}^\infty n\,c_n x^{n-1} \qquad\text{and}\qquad \int f(x)\,dx = C + \sum_{n=0}^\infty \frac{c_n}{n+1}x^{n+1}. \]
The radius stays the same. The endpoints might change, so recheck them if you need the interval.
The key example. Differentiate both sides of \(\frac{1}{1-x} = \sum x^n\):
\[ \frac{1}{(1-x)^2} = \sum_{n=1}^\infty n\,x^{n-1} = 1 + 2x + 3x^2 + 4x^3 + \cdots \]
The \(n = 0\) term was a constant, so its derivative is \(0\) and the sum now starts at \(n = 1\).
Reindexing so the exponent is \(n\). WeBWorK usually wants \(\sum_{n=0}^\infty\). Substitute \(m = n - 1\), so \(n = m + 1\), then rename \(m\) back to \(n\):
\[ \sum_{n=1}^\infty n\,x^{n-1} = \sum_{n=0}^\infty (n+1)\,x^n \]
Check that the first terms agree: \(1, 2x, 3x^2, \ldots\) in both. ✓
Example (chain rule inside). \(\displaystyle\frac{2x}{(1-x^2)^2}\). Recognize it as a derivative: \(\frac{d}{dx}\frac{1}{1-x^2} = \frac{2x}{(1-x^2)^2}\). So differentiate \(\frac{1}{1-x^2} = \sum x^{2n}\):
\[ \frac{2x}{(1-x^2)^2} = \sum_{n=1}^\infty 2n\,x^{2n-1} = \sum_{n=0}^\infty 2(n+1)\,x^{2n+1}, \qquad |x| < 1 \]
\(\dfrac{(\text{inside})'}{(1 - \text{inside})^2}\) is \(\dfrac{d}{dx}\dfrac{1}{1-\text{inside}}\). Look for the derivative of the inside sitting in the numerator.
5. Integrating term by term (#6, 9)
Example: \(\ln(1+x)\). Since \(\frac{d}{dx}\ln(1+x) = \frac{1}{1+x} = \sum (-1)^n x^n\),
\[ \ln(1+x) = C + \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n+1} \]
Find \(C\) by plugging in the center. At \(x = 0\), the left side is \(\ln 1 = 0\) and every term on the right is \(0\), so \(C = 0\):
\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n+1}, \qquad |x| < 1 \]
Example: \(\arctan x\). Integrate \(\frac{1}{1+x^2} = \sum(-1)^n x^{2n}\) (again \(C = \arctan 0 = 0\)):
\[ \arctan x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots \]
Example where \(C \ne 0\): \(\ln(2 + x)\). The derivative is \(\frac{1}{2+x} = \frac12\cdot\frac{1}{1+\frac x2} = \sum (-1)^n\frac{x^n}{2^{n+1}}\). Integrate:
\[ \ln(2+x) = C + \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{(n+1)\,2^{n+1}} \]
At \(x = 0\): \(\ln 2 = C + 0\), so \(C = \ln 2\):
\[ \ln(2+x) = \ln 2 + \sum_{n=0}^\infty \frac{(-1)^n}{n+1}\left(\frac{x}{2}\right)^{n+1}, \qquad |x| < 2 \]
For \(\ln(a \pm x)\) with \(a \ne 1\), the constant is \(\ln a\). For \(\ln(a - x)\), watch the sign too: \(\frac{d}{dx}\ln(a - x) = \frac{-1}{a - x}\).
Example: integrate and match a known function (like #6). Start from the series
\[ -\frac{1}{(1+x)^2} = \sum_{n=1}^\infty (-1)^n\, n\, x^{n-1} \]
(it’s the derivative of \(\frac{1}{1+x} = \sum(-1)^n x^n\)). Integrating gives \(\displaystyle\int -\frac{1}{(1+x)^2}\,dx = \frac{1}{1+x} + C\) on the function side, and
\[ \sum_{n=1}^\infty (-1)^n x^n + C = -x + x^2 - x^3 + \cdots + C \]
on the series side. To make it equal \(\frac{1}{1+x} = 1 - x + x^2 - \cdots\), take \(C = 1\), the missing \(n = 0\) term. So the answer is \(\sum_{n=0}^\infty(-1)^n x^n\).
6. Evaluating numerical sums with \(f'\) and \(f''\) (#7, 8)
Sums like \(\displaystyle\sum\frac{n}{4^n}\) aren’t geometric, but they are a power series evaluated at a number:
\[ \sum_{n=1}^\infty \frac{n}{4^n} = \sum_{n=1}^\infty n\,x^n\,\Big|_{x = 1/4} \]
From §4, \(\sum_{n=1}^\infty n\,x^{n-1} = \frac{1}{(1-x)^2}\). Multiply by \(x\) to fix the exponent:
\[ \sum_{n=1}^\infty n\,x^n = x\cdot f'(x) = \frac{x}{(1-x)^2} \]
At \(x = \frac14\):
\[ \sum_{n=1}^\infty \frac{n}{4^n} = \frac{1/4}{(3/4)^2} = \frac14\cdot\frac{16}{9} = \frac49 \]
Second derivative. Differentiating again gives \(f''(x) = \frac{2}{(1-x)^3} = \sum_{n=2}^\infty n(n-1)x^{n-2}\), so
\[ \sum_{n=2}^\infty n(n-1)x^n = x^2 f''(x) = \frac{2x^2}{(1-x)^3} \]
At \(x = \frac14\):
\[ \sum_{n=2}^\infty\frac{n(n-1)}{4^n} = \frac{2\cdot\frac{1}{16}}{\left(\frac34\right)^3} = \frac18\cdot\frac{64}{27} = \frac{8}{27} \]
Bonus. Since \(n^2 = n(n-1) + n\), we get \(\displaystyle\sum_{n=1}^\infty\frac{n^2}{4^n} = \frac{8}{27} + \frac49 = \frac{20}{27}\).
\(f'(\frac14) = \sum n\left(\frac14\right)^{n-1}\), which has exponent \(n - 1\). The sum \(\sum\frac{n}{4^n}\) has exponent \(n\). So you need \(x f'(x)\) at \(x = \frac14\), which is \(\frac14 f'(\frac14)\). That’s why the WeBWorK answers look like \(\frac13\cdot(\text{number})\) and \(\frac{1}{3^2}\cdot(\text{number})\).
Summary checklist
| Problem type | HW 6.2 | Key move |
|---|---|---|
| Partial fractions → series | #1 | split, factor constants to get \(\frac{1}{1-u}\), combine into one \(\sum\); check \(f(0)\) |
| Series → rational function | #2, 3 | first term over $1 - $ ratio; group repeating blocks, giving \(\frac{p(x)}{1-x^m}\) |
| Product of series | #4 | \(c_n = \sum_{k=0}^n a_k b_{n-k}\) |
| Differentiate | #5, 10 | \(\frac{1}{(1-x)^2} = \sum (n+1)x^n\); reindex to start at \(n=0\) |
| Integrate | #6, 9 | add \(+C\), find it by plugging in the center |
| Numerical sums | #7, 8 | \(\sum n x^n = x f'(x)\), \(\sum n(n-1)x^n = x^2 f''(x)\) |