§1 · Radius of Convergence
Quick Memory Key
RATIO & ROOT — Two tests that always find the radius. Use Ratio Test when you see factorials \(n!\) or \(a^n\). Use Root Test when you see \(n\)-th powers like \((a_n)^n\). Both give \(R = 1/L\).
Q1
Radius
★☆☆ · Ratio Test
Find the radius of convergence \(R\) of:
\[\sum_{n=0}^{\infty} \frac{x^n}{n!}\]
\[\sum_{n=0}^{\infty} \frac{x^n}{n!}\]
💡 FACTORIAL → RATIO TEST
💬 Explanation
Apply the Ratio Test: compute \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\) where \(a_n = x^n/n!\).
\[\left|\frac{x^{n+1}/(n+1)!}{x^n/n!}\right| = \frac{|x|}{n+1} \to 0 \text{ as } n\to\infty\]
Since \(L=0 < 1\) for all \(x\), the series converges for every \(x \in \mathbb{R}\).
Therefore \(R = \infty\).
Q2
Radius
★☆☆ · Ratio Test
Find \(R\) for \(\displaystyle\sum_{n=1}^{\infty} n!\, x^n\).
💡 GROWS FAST → TINY R
💬 Explanation
Ratio Test: \(\left|\frac{(n+1)!\,x^{n+1}}{n!\,x^n}\right| = (n+1)|x| \to \infty\) unless \(x=0\).
The series diverges for all \(x \neq 0\), so \(R = 0\).
Q3
Radius
★★☆ · Root Test
Find \(R\) for \(\displaystyle\sum_{n=1}^{\infty} \left(\frac{x}{3}\right)^n\).
💡 GEOMETRIC: R = base denominator
💬 Explanation
This is a geometric series with ratio \(r = x/3\). Converges when \(|r| < 1\).
\(|x/3| < 1 \Rightarrow |x| < 3\). So \(R = 3\).
Q4
Radius
★★☆ · Ratio Test (Tricky)
Find \(R\) for \(\displaystyle\sum_{n=0}^{\infty} \frac{(-3)^n\, x^n}{\sqrt{n+1}}\).
💡 IGNORE SIGN → |coefficient|^(1/n)
💬 Explanation
Ratio Test with \(a_n = (-3)^n x^n/\sqrt{n+1}\):
\[\left|\frac{a_{n+1}}{a_n}\right| = 3|x|\cdot\sqrt{\frac{n+1}{n+2}} \to 3|x|\]
Converges when \(3|x| < 1\), i.e. \(|x| < 1/3\). So \(R = 1/3\).
§2 · Interval of Convergence
Quick Memory Key
ENDPOINT CHECK — After finding \(R\), always plug in \(x = c \pm R\) separately. At endpoints, use p-series, alternating series, or divergence test. Most mistakes happen here!
Q5
Interval
★★☆ · Endpoint Check
The series \(\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}\) has \(R=1\).
At \(x = 1\): does it converge or diverge?
Type:
At \(x = 1\): does it converge or diverge?
Type:
converge or diverge
💡 HARMONIC SERIES = DIVERGE
💬 Explanation
At \(x=1\): \(\sum 1/n\) — this is the harmonic series, which diverges (p-series with \(p=1 \leq 1\)).
\(\sum_{n=1}^{\infty} \frac{1}{n} = \infty\) → DIVERGE
Q6
Interval
★★☆ · Alternating Series
For \(\displaystyle\sum_{n=1}^{\infty}\frac{x^n}{n}\) (same series), at \(x = -1\): converge or diverge?
💡 ALTERNATING + DECREASING → CONVERGE
💬 Explanation
At \(x=-1\): \(\sum (-1)^n/n\) — Alternating Harmonic Series.
By the Alternating Series Test: \(1/n\) decreases to 0 → CONVERGES.
In fact it converges to \(-\ln 2\).
In fact it converges to \(-\ln 2\).
Q7
Interval
★★★ · Full Interval
What is the interval of convergence of \(\displaystyle\sum_{n=0}^{\infty}\frac{(x-2)^n}{n\cdot 3^n}\)?
Type the interval, e.g.
Type the interval, e.g.
-1 to 5
💡 CENTER = 2, SHIFT YOUR INTERVAL
Hint — Radius first
Ratio Test on \(\frac{(x-2)^n}{n \cdot 3^n}\): let \(u = x-2\), find \(R\), then check endpoints \(u = \pm R\).
💬 Explanation
Let \(u = x-2\). Ratio Test gives \(L = |u|/3\). So \(R=3\), center \(c=2\).
Endpoints: \(x = -1\) gives \(\sum\frac{(-1)^n}{n}\) → converges (alternating).
\(x = 5\) gives \(\sum\frac{1}{n}\) → diverges (harmonic).
Interval: \([-1,\, 5)\) or -1 to 5
\(x = 5\) gives \(\sum\frac{1}{n}\) → diverges (harmonic).
Interval: \([-1,\, 5)\) or -1 to 5
§3 · Convergence Tests
Quick Memory Key
PICK A TEST FLOWCHART — (1) Divergence Test first. (2) Geometric / p-series? Use formula. (3) Factorial? Ratio. (4) \(n\)-th power? Root. (5) Comparison / Limit Comparison for everything else.
Q8
Convergence
★☆☆ · p-Series
Does \(\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}\) converge or diverge?
💡 p-SERIES: p>1 → CONVERGE
💬 Explanation
p-series \(\sum 1/n^p\) converges if and only if \(p > 1\).
Here \(p = 2 > 1\) → CONVERGE. (Equals \(\pi^2/6\))
Q9
Convergence
★★☆ · Ratio Test Confusion
The Ratio Test gives \(L = 1\) for a series. What can you conclude?
Type:
Type:
converge, diverge, or inconclusive
💡 L=1 → TEST FAILS, try another!
💬 Explanation
The Ratio (and Root) Test is inconclusive when \(L = 1\). Use a different test.
\(L < 1\) → Converge · \(L > 1\) → Diverge · \(L = 1\) → INCONCLUSIVE
Q10
Convergence
★★☆ · Root Test
Find \(R\) for \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{2n+1}{3n-1}\right)^n x^n\).
💡 (...)^n → ROOT TEST instantly
💬 Explanation
Root Test: \(\limsup_{n\to\infty}|a_n|^{1/n} = \lim_{n\to\infty}\frac{2n+1}{3n-1}\cdot|x|\).
\[\lim_{n\to\infty}\frac{2n+1}{3n-1} = \frac{2}{3}\]
Converges when \(\frac{2}{3}|x| < 1\), i.e. \(|x| < \frac{3}{2}\). So \(R = \dfrac{3}{2}\).
§4 · Taylor & Maclaurin Series
Quick Memory Key
BIG 5 MACLAURIN — Memorize these cold:
eˣ: \(\sum x^n/n!\) · sin x: odd terms · cos x: even terms · ln(1+x): \(\sum(-1)^{n+1}x^n/n\) · 1/(1-x): \(\sum x^n\)
Q11
Maclaurin
★☆☆ · Standard Formula
What is the Maclaurin series for \(e^x\)?
Write the general term only: type
Write the general term only: type
x^n/n!
💡 e^x = sum of ALL x^n/n!
💬 Explanation
For \(e^x\), all derivatives equal \(e^x\), so \(f^{(n)}(0) = 1\) for all \(n\).
\[e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots\]
General term: x^n/n!
Q12
Maclaurin
★★☆ · Substitution Trick
Using the known series for \(e^x\), write the Maclaurin series for \(e^{-x^2}\).
Type the general term: e.g.
Type the general term: e.g.
(-1)^n x^(2n)/n!
💡 SUBSTITUTE x → -x^2 directly
💬 Explanation
Replace \(x\) with \(-x^2\) in the series for \(e^x\):
\[e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!} = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{n!}\]
General term: (-1)^n x^(2n)/n!
Q13
Taylor
★★☆ · Taylor at a = 1
Find the 3rd-degree Taylor polynomial \(T_3(x)\) for \(f(x) = \ln x\) centered at \(a = 1\).
What is the coefficient of \((x-1)^3\)?
What is the coefficient of \((x-1)^3\)?
💡 COEFFICIENT = f'''(a)/3!
💬 Explanation
Compute derivatives at \(a=1\): \(f'''(x) = 2/x^3\), so \(f'''(1) = 2\).
Coefficient of \((x-1)^3\) = \(\dfrac{f'''(1)}{3!} = \dfrac{2}{6} = \dfrac{1}{3}\)
Answer: 1/3
Answer: 1/3
Q14
Maclaurin
★★☆ · sin x Series
The Maclaurin series for \(\sin x\) only has odd powers. What is the general term?
Type:
Type:
(-1)^n x^(2n+1)/(2n+1)!
💡 SIN = ODD POWERS, alternating signs
💬 Explanation
\(\sin x = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \cdots\)
\[\sin x = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}\]
General term: (-1)^n x^(2n+1)/(2n+1)!
Q15
Maclaurin
★★☆ · Differentiation of Series
Start from \(\dfrac{1}{1-x} = \sum_{n=0}^{\infty}x^n\).
Differentiate both sides to find a series for \(\dfrac{1}{(1-x)^2}\).
What is the general term?
Differentiate both sides to find a series for \(\dfrac{1}{(1-x)^2}\).
What is the general term?
💡 DIFFERENTIATE TERM BY TERM
💬 Explanation
Differentiate \(\sum x^n\) term by term: \(\sum nx^{n-1}\). Reindex by letting \(m = n-1\):
\[\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n\,x^{n-1} = \sum_{n=0}^{\infty}(n+1)x^n\]
General term: (n+1)x^n
§5 · Taylor's Remainder & Error Bound
Quick Memory Key
LAGRANGE REMAINDER — Error bound formula: \(|R_n(x)| \leq \dfrac{M}{(n+1)!}|x-a|^{n+1}\) where \(M\) is the max of \(|f^{(n+1)}|\) on the interval. For alternating series, error \(\leq\) first omitted term.
Q16
Error
★★★ · Alternating Error Bound
The series \(\cos x = 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots\)
is approximated by \(T_2(x) = 1 - \dfrac{x^2}{2}\) at \(x = 0.1\).
The error \(|R_2| \leq\) first omitted term. What is that bound?
Type as a fraction or decimal, e.g.
is approximated by \(T_2(x) = 1 - \dfrac{x^2}{2}\) at \(x = 0.1\).
The error \(|R_2| \leq\) first omitted term. What is that bound?
Type as a fraction or decimal, e.g.
1/24 or just the term
💡 ALTERNATING ERROR = NEXT TERM VALUE
💬 Explanation
The next term after \(T_2\) is \(-x^4/4!\). At \(x = 0.1\):
\[\left|\frac{(0.1)^4}{4!}\right| = \frac{0.0001}{24} \approx 4.17\times10^{-6}\]
Error bound \(\leq \dfrac{(0.1)^4}{24} = \dfrac{1}{2{,}400{,}000}\). Accept: 1/24 (unsubstituted form is fine too).
Q17
Error
★★★ · Lagrange Remainder
For \(f(x)=e^x\) centered at \(a=0\), the Lagrange remainder after \(n\) terms is:
\[|R_n(x)| \leq \frac{e^c}{(n+1)!}|x|^{n+1}\] where \(0 \leq c \leq x\). If \(x = 1\), what value do we use as an upper bound for \(e^c\)?
Type a number, e.g.
\[|R_n(x)| \leq \frac{e^c}{(n+1)!}|x|^{n+1}\] where \(0 \leq c \leq x\). If \(x = 1\), what value do we use as an upper bound for \(e^c\)?
Type a number, e.g.
e or 3
💡 e^c IS BOUNDED BY e when x=1
💬 Explanation
Since \(0 \leq c \leq 1\), we have \(e^c \leq e^1 = e\).
Upper bound for \(e^c\) is simply \(e \approx 2.718\).
So \(|R_n(1)| \leq \dfrac{e}{(n+1)!}\). Answer: e
So \(|R_n(1)| \leq \dfrac{e}{(n+1)!}\). Answer: e
§6 · Applications & Operations
Quick Memory Key
INTEGRATE · DIFFERENTIATE · SUBSTITUTE — Power series can be integrated and differentiated term-by-term inside \((-R,R)\). These operations preserve the radius (not necessarily endpoints). Substitution is the fastest shortcut for new functions.
Q18
Operations
★★☆ · Integration of Series
Use \(\dfrac{1}{1+x^2} = \sum_{n=0}^{\infty}(-1)^n x^{2n}\) to derive a series for \(\arctan x\).
What is the general term after integration?
What is the general term after integration?
💡 INTEGRATE 1/(1+x^2) = arctan x
💬 Explanation
Integrate \(\sum(-1)^n x^{2n}\) term by term:
\[\arctan x = \int_0^x \frac{1}{1+t^2}\,dt = \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1}\]
General term: (-1)^n x^(2n+1)/(2n+1)
Q19
Taylor
★★★ · Classic Limit via Series
Evaluate: \[\lim_{x \to 0} \frac{1 - \cos x}{x^2}\]
using the Maclaurin series for \(\cos x\).
💡 SERIES → CANCEL → TRIVIAL LIMIT
💬 Explanation
Substitute \(\cos x = 1 - x^2/2 + x^4/24 - \cdots\):
\[\frac{1-\cos x}{x^2} = \frac{x^2/2 - x^4/24 + \cdots}{x^2} = \frac{1}{2} - \frac{x^2}{24} + \cdots\]
As \(x\to 0\), this \(\to \dfrac{1}{2}\). Answer: 1/2
Q20
Taylor
★★★ · Binomial Series
The binomial series states \((1+x)^k = \sum_{n=0}^{\infty}\binom{k}{n}x^n\) for \(|x|<1\).
Using this, the Maclaurin series for \(\dfrac{1}{\sqrt{1-x}}\) has general term:
Type:
Using this, the Maclaurin series for \(\dfrac{1}{\sqrt{1-x}}\) has general term:
Type:
(2n)! x^n / (4^n (n!)^2)
💡 BINOMIAL k=-1/2, SUBSTITUTE -x
Hint
Write \((1-x)^{-1/2}\) as \((1+(-x))^{-1/2}\), then apply the binomial series with \(k = -\tfrac{1}{2}\).
💬 Explanation
With \(k = -1/2\) and \(x \to -x\):
\[\binom{-1/2}{n}(-x)^n = \frac{(2n)!}{4^n(n!)^2}x^n\]
\[\frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty}\frac{(2n)!}{4^n(n!)^2}x^n\]
General term: (2n)! x^n / (4^n (n!)^2)