Power Series

The general form is \(\sum_{n=0}^{\infty} a_n(x-c)^n\) where \(c\) is the center. Option A is the special case \(c=0\) (centered at origin). Option C is a specific series (\(e^x\) without the center shift). Option D is not a power series at all — the power of \(x\) never changes.

The center \(c\) is the value subtracted from \(x\) inside the parentheses. Here \((x - 2)^n\) means \(c = 2\). The coefficient \(3^n\) is just \(a_n\) — it doesn't affect the center. Common trap: confusing coefficients with the center!

Apply Ratio Test: \(\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right| = \left|\frac{x}{n+1}\right| \to 0\) as \(n \to \infty\), for ANY fixed \(x\). Since \(L = 0 < 1\) always, the series converges for all \(x\). Therefore \(R = \infty\). (This is \(e^x\)!)

Ratio Test: \(\left|\frac{(n+1)!\,x^{n+1}}{n!\,x^n}\right| = (n+1)|x| \to \infty\) for any \(x \neq 0\). Since \(L = \infty > 1\) for all \(x \neq 0\), the series only converges at \(x = 0\). So \(R = 0\). Key insight: factorials in the numerator make \(R = 0\); factorials in the denominator make \(R = \infty\).

With \(a_n = \frac{1}{n \cdot 4^n}\): Ratio Test gives \(\frac{a_{n+1}}{a_n} = \frac{n}{n+1} \cdot \frac{1}{4} \to \frac{1}{4}\). So \(L = \frac{|x-3|}{4}\), converging when \(|x-3| < 4\). The radius is \(R = 4\). The center \(c=3\) does NOT affect the radius!

Let \(u = x^2\). The series \(\sum a_n x^{2n} = \sum a_n u^n\) converges when \(|u| < 5\), i.e., \(|x^2| < 5\), i.e., \(|x| < \sqrt{5}\). So the new radius is \(R = \sqrt{5}\).

Rule: replacing \(x^n\) with \(x^{2n}\) takes the square root of the radius.

Ratio Test → \(R = 1\). Open interval: \((-1, 1)\).

• At \(x = 1\): \(\sum \frac{1}{n}\) = harmonic series → diverges.
• At \(x = -1\): \(\sum \frac{(-1)^n}{n}\) = alternating harmonic → converges (AST).

So include \(-1\) but not \(1\): interval is \([-1,\, 1)\).

Center: \(c = -1\), \(R = 3\). Open interval: \((-4, 2)\).

• At \(x = 2\): \(\left(\frac{3}{3}\right)^n = 1^n\) → \(\sum 1\) → diverges.
• At \(x = -4\): \(\left(\frac{-3}{3}\right)^n = (-1)^n\) → \(\sum (-1)^n\) → diverges.

Both endpoints fail → interval is \((-4,\, 2)\).

The series is \(\sum \frac{(-1)^n x^n}{n^2}\). At both endpoints \(|x|=1\), the absolute values give \(\sum \frac{1}{n^2}\), which is a convergent \(p\)-series (\(p=2>1\)). So both endpoints converge absolutely, giving interval \([-1, 1]\). Beware: the \((-1)^n\) in the coefficient is a red herring at endpoints — absolute convergence covers both.

Replace \(x\) with \(-x\) in \(\frac{1}{1-x} = \sum x^n\): we get \(\frac{1}{1+x} = \sum (-x)^n = \sum (-1)^n x^n\). Option A is \(\frac{1}{1-x}\). Option C is \(\sum x^{2n} = \frac{1}{1-x^2}\). Option D starts at \(n=1\) and skips the constant term.

Cosine has only even powers: \(\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\). Option A is \(\sin x\) (odd powers). Option C is \(e^x\) (all terms positive). Option D is \(e^{-x}\) (alternating all powers).

Differentiate term-by-term: \(\frac{d}{dx}\left[\frac{x^{n+1}}{n+1}\right] = x^n\). So \(f'(x) = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}\). Option B has a division by \(n\), which is wrong for \(n=0\). Option C starts at \(n=1\), missing the constant term. Option D would come from differentiating \(x^n\), not \(\frac{x^{n+1}}{n+1}\).

\(\frac{1}{1+x} = \sum (-1)^n x^n\). Integrate: \(\ln(1+x) = \sum_{n=0}^\infty \frac{(-1)^n x^{n+1}}{n+1}\). Re-indexing with \(m = n+1\): \(= \sum_{m=1}^\infty \frac{(-1)^{m-1} x^m}{m} = \sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n}\). Options B and C and D are equivalent forms — D and C are both correct! A is missing the alternating sign (that would be \(-\ln(1-x)\)).

The theorem states: term-by-term differentiation preserves the radius \(R\) exactly. However, convergence at the endpoints \(x = c \pm R\) must be re-examined for the new differentiated series. Example: \(\sum \frac{x^n}{n}\) converges at \(x=-1\), but after differentiating to \(\sum x^{n-1}\), neither endpoint converges.

Method 1 (substitute): \(e^{2x} = \sum \frac{(2x)^n}{n!} = \sum \frac{2^n x^n}{n!}\). For \(n=3\): coefficient \(= \frac{2^3}{3!} = \frac{8}{6} = \frac{4}{3}\). Options A and B miss the factor of \(2^3\). Option D equals \(\frac{4}{3}\) too — but \(\frac{8}{6}\) is the unsimplified form; C is the simplified correct answer \(\frac{4}{3}\).

For \(\sin x\), only odd-power terms appear: \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\). The third-degree polynomial keeps up to \(x^3\): \(T_3(x) = x - \frac{x^3}{6}\). Option A has an \(x^2\) term — that belongs to \(\cos x\). Option C is the second-degree Taylor of \(\cos x\). Option D has the wrong sign.

Substitute \(\sin x = x - \frac{x^3}{6} + O(x^5)\):

\(\frac{\sin x - x}{x^3} = \frac{-\frac{x^3}{6} + O(x^5)}{x^3} = -\frac{1}{6} + O(x^2) \to -\frac{1}{6}\)

This is one of the most powerful applications of series: turn \(\frac{0}{0}\) limits into simple algebra!

Integrate \(\frac{1}{1+x^2} = \sum_{n=0}^\infty (-1)^n x^{2n}\) term-by-term: \(\int \frac{dx}{1+x^2} = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1} = \arctan x\).

Check: \(\sin x\) has factorials in denominator. \(\tanh^{-1} x\) has no alternating sign. \(\ln(1+x)\) has linear powers only.

\(e^{-x^2} = 1 - x^2 + \frac{x^4}{2} - \cdots\). Using first two terms: \(\int_0^{0.1}(1-x^2)\,dx = \left[x - \frac{x^3}{3}\right]_0^{0.1} = 0.1 - \frac{0.001}{3} \approx 0.1 - 0.000\overline{3} \approx 0.09967\).

The exact value is ≈ 0.09983 (three terms gives this). Option A ignores the \(x^2\) correction entirely.

The series \(\sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\) is exactly the Maclaurin series of \(\cos x\).

Compare: \(\cosh x = \sum \frac{x^{2n}}{(2n)!}\) — same structure but NO alternating sign. \(\sin x\) uses odd powers \(x^{2n+1}\). \(e^{-x} = \sum \frac{(-x)^n}{n!}\) alternates all powers, not just even ones.