Sequences and Series — Diagnostic Tests [BC Only]

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for sequences and series.

UT-1: Choosing the Correct Convergence Test for a Tricky Series

Question:

Determine whether $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n} + (-1)^n}$ converges conditionally, converges absolutely, or diverges.

A student applies the alternating series test and concludes it converges conditionally because $\dfrac{1}{\sqrt{n} + (-1)^n}$ decreases to $0$. Identify the flaw in this reasoning and determine the correct answer.

Solution:

The alternating series test requires that $b_n = \dfrac{1}{\sqrt{n} + (-1)^n}$ is positive and decreasing. Let us check if $b_n$ is decreasing.

$b_1 = \dfrac{1}{1 - 1}$ is undefined (division by zero). So the series is not even well-defined starting from $n = 1$.

Start from $n = 2$ instead: b_2 = \dfrac{1}{\sqrt{2} + 1} \approx 0.414$$b_3 = \dfrac{1}{\sqrt{3} - 1} \approx 1.366$$b_4 = \dfrac{1}{2 + 1} \approx 0.333.

Since $b_3 > b_2$The sequence $\{b_n\}$ is not decreasing, so the alternating series test does not apply.

To determine convergence, rewrite:

$\frac{(-1)^n}{\sqrt{n} + (-1)^n} = \frac{(-1)^n(\sqrt{n} - (-1)^n)}{n - 1} = \frac{(-1)^n\sqrt{n}}{n-1} - \frac{1}{n-1}$

$= \frac{(-1)^n}{\sqrt{n}} \cdot \frac{n}{n-1} - \frac{1}{n-1}$

The first part $\dfrac{(-1)^n}{\sqrt{n}} \cdot \dfrac{n}{n-1}$ converges by the alternating series test (since $\dfrac{n}{\sqrt{n}(n-1)} = \dfrac{\sqrt{n}}{n-1} \to 0$ and is eventually decreasing).

The second part $-\dfrac{1}{n-1} = -\displaystyle\sum_{n=2}^{\infty}\dfrac{1}{n-1} = -\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k}$ is the negative harmonic series, which diverges.

Since a convergent series minus a divergent series diverges, the original series diverges.

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UT-2: Taylor Series Remainder and Lagrange Error Bound

Question:

Use the Maclaurin series for $e^x$ to approximate $e^{0.3}$ using the first three nonzero terms.

(a) Compute the approximation. (b) Use the Lagrange error bound to find an upper bound on the absolute error. (c) The actual value is $e^{0.3} \approx 1.3498588$. Compute the actual error and verify it is within the bound. (d) A student claims “since the Maclaurin series for $e^x$ converges for all $x$The error must go to zero.” Explain why this does not mean the error is zero for any finite number of terms.

Solution:

(a) The Maclaurin series: $e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \cdots$

First three nonzero terms: $1 + 0.3 + \dfrac{0.09}{2} = 1 + 0.3 + 0.045 = 1.345$.

(b) The Lagrange remainder after $n$ terms: $|R_n(x)| \leq \dfrac{M|x|^{n+1}}{(n+1)!}$ where $M = \max|f^{(n+1)}(c)|$ for $c$ between $0$ and $x$.

After 3 terms ($n = 3$Using up to the $x^2$ term), the next term involves $f^{(3)}(x) = e^x$:

$|R_2(0.3)| \leq \frac{e^{0.3} \cdot (0.3)^3}{3!} = \frac{e^{0.3} \cdot 0.027}{6}$

Since $e^{0.3} \lt e^1 \lt 3$:

$|R_2(0.3)| \lt \frac{3 \cdot 0.027}{6} = \frac{0.081}{6} = 0.0135$

A tighter bound using $e^{0.3} \lt 1.35$:

$|R_2(0.3)| \lt \frac{1.35 \cdot 0.027}{6} = 0.006075$

(c) Actual error: $|e^{0.3} - 1.345| = |1.3498588 - 1.345| = 0.0048588$.

Is $0.0048588 \lt 0.0135$? Yes. Is $0.0048588 \lt 0.006075$? Yes. The error is within both bounds.

(d) Convergence of the series means the partial sums approach $e^x$ as the number of terms goes to infinity. For any finite number of terms, there is a nonzero remainder. The series “converging” is a statement about the limit of partial sums, not about any individual partial sum. This is the distinction between an infinite process and its finite approximation.

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UT-3: Radius and Interval of Convergence with Endpoint Analysis

Question:

Find the radius and interval of convergence of $\displaystyle\sum_{n=1}^{\infty} \frac{(x - 2)^n}{n \cdot 3^n}$.

Solution:

Apply the ratio test:

$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{(x-2)^{n+1}}{(n+1) \cdot 3^{n+1}} \cdot \frac{n \cdot 3^n}{(x-2)^n}\right| = \lim_{n \to \infty}\frac{n}{n+1} \cdot \frac{|x-2|}{3} = \frac{|x-2|}{3}$

The ratio test gives convergence when $\dfrac{|x-2|}{3} \lt 1$I.e., $|x - 2| \lt 3$.

Radius of convergence: $R = 3$.

Endpoints:

$x = 5$: $\displaystyle\sum_{n=1}^{\infty} \frac{3^n}{n \cdot 3^n} = \sum_{n=1}^{\infty}\frac{1}{n}$ — the harmonic series, which diverges.

$x = -1$: $\displaystyle\sum_{n=1}^{\infty} \frac{(-3)^n}{n \cdot 3^n} = \sum_{n=1}^{\infty}\frac{(-1)^n}{n}$ — the alternating harmonic series, which converges conditionally (by the alternating series test: $\dfrac{1}{n} \to 0$ and decreases).

Interval of convergence: $[-1, 5)$.

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Integration Tests

Tests synthesis of sequences and series with other topics.

IT-1: Power Series Differentiation to Evaluate a Sum (with Derivatives)

Question:

Starting from the geometric series $\displaystyle\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ for $|x| \lt 1$Find the exact value of $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{2^n}$.

Solution:

From $\displaystyle\sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$Differentiate both sides:

$\sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2}$

Multiply by $x$:

$\sum_{n=1}^{\infty} nx^n = \frac{x}{(1-x)^2}$

Differentiate again:

$\sum_{n=1}^{\infty} n^2 x^{n-1} = \frac{(1-x)^2 + 2x(1-x)}{(1-x)^4} = \frac{(1-x)(1-x + 2x)}{(1-x)^4} = \frac{1+x}{(1-x)^3}$

Multiply by $x$:

$\sum_{n=1}^{\infty} n^2 x^n = \frac{x(1+x)}{(1-x)^3}$

Set $x = \dfrac{1}{2}$:

$\sum_{n=1}^{\infty}\frac{n^2}{2^n} = \frac{\frac{1}{2} \cdot \frac{3}{2}}{\left(\frac{1}{2}\right)^3} = \frac{\frac{3}{4}}{\frac{1}{8}} = 6$

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IT-2: Taylor Polynomial Integration (with Integrals)

Question:

(a) Find the fourth-degree Maclaurin polynomial $T_4(x)$ for $f(x) = \cos(x^2)$. (b) Use $T_4(x)$ to approximate $\displaystyle\int_0^{0.5} \cos(x^2)\,dx$. (c) Bound the error in this approximation using the Lagrange remainder.

Solution:

(a) Start with $\cos u = 1 - \dfrac{u^2}{2!} + \dfrac{u^4}{4!} - \cdots$. Substitute $u = x^2$:

$\cos(x^2) = 1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \cdots = 1 - \frac{x^4}{2} + \frac{x^8}{24} - \cdots$

The fourth-degree Maclaurin polynomial (all terms through $x^4$):

$T_4(x) = 1 - \frac{x^4}{2}$

(b) $\displaystyle\int_0^{0.5} T_4(x)\,dx = \int_0^{0.5}\left(1 - \frac{x^4}{2}\right)dx = \left[x - \frac{x^5}{10}\right]_0^{0.5} = 0.5 - \frac{1}{320} = \frac{160 - 1}{320} = \frac{159}{320} \approx 0.496875$

(c) The next term in the series is $\dfrac{x^8}{24}$. The error from truncating after the $x^4$ term is bounded by:

$|R_4(x)| \leq \frac{M|x|^6}{6!}$

Where $M = \max|f^{(6)}(c)|$ for $c \in [0, 0.5]$. This is complicated. Instead, bound using the next series term:

Since the series for $\cos(x^2)$ is alternating and the terms decrease in magnitude for $|x| \lt 1$The error in truncating after the $x^4$ term is at most the magnitude of the next term:

$|R_4(x)| \leq \frac{x^8}{24}$

So the error in the integral is bounded by:

$\left|\int_0^{0.5} R_4(x)\,dx\right| \leq \int_0^{0.5}\frac{x^8}{24}\,dx = \frac{1}{24}\cdot\frac{(0.5)^9}{9} = \frac{1}{24 \cdot 9 \cdot 512} = \frac{1}{110592} \approx 0.00000904$

More rigorously, using the Lagrange form: $f^{(5)}(x)$ involves $\sin(x^2)$ and $\cos(x^2)$ terms. The maximum of $|f^{(5)}(c)|$ on $[0, 0.5]$ is bounded (all derivatives of $\cos(x^2)$ are bounded by polynomials in $c$ times sines and cosines). The alternating series bound gives a cleaner result and is sufficient for the AP exam.

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IT-3: Series Convergence via Integral Test (with Integrals and Limits)

Question:

Determine whether $\displaystyle\sum_{n=2}^{\infty} \frac{1}{n(\ln n)^p}$ converges for: (a) $p = 2$ (b) $p = 1$ (c) General $p$

Then, for the convergent case(s), use the integral test remainder bound to determine how many terms are needed to approximate the sum to within $0.001$.

Solution:

Let $f(x) = \dfrac{1}{x(\ln x)^p}$ for $x \geq 2$. This is positive, continuous, and decreasing for $x \geq 2$ (when $p > 0$).

Let $u = \ln x$, $du = \dfrac{dx}{x}$:

$\int_2^{\infty}\frac{dx}{x(\ln x)^p} = \int_{\ln 2}^{\infty}\frac{du}{u^p}$

(a) $p = 2$: $\displaystyle\int_{\ln 2}^{\infty}\frac{du}{u^2} = \left[-\frac{1}{u}\right]_{\ln 2}^{\infty} = \frac{1}{\ln 2} \lt \infty$. The series converges.

(b) $p = 1$: $\displaystyle\int_{\ln 2}^{\infty}\frac{du}{u} = \left[\ln u\right]_{\ln 2}^{\infty} = \infty$. The series diverges.

(c) The $p$-integral $\displaystyle\int \frac{du}{u^p}$ converges iff $p > 1$. Therefore:

\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^p} \text{ converges iff p > 1

For the remainder bound with $p = 2$: the error from using $N$ terms satisfies:

$R_N \leq \int_N^{\infty}\frac{dx}{x(\ln x)^2} = \frac{1}{\ln N}$

We need $\dfrac{1}{\ln N} \lt 0.001$:

$\ln N > 1000 \implies N > e^{1000}$

This is astronomically large, showing that despite convergence, the series converges extremely slowly. This illustrates an important limitation of the integral test for error bounds: convergence does not imply practical computability.

For comparison, even $p = 1.01$ gives:

$\int_N^{\infty}\frac{du}{u^{1.01}} = \frac{N^{-0.01}}{0.01} = 100N^{-0.01}$

Setting $100N^{-0.01} \lt 0.001$: $N^{-0.01} \lt 0.00001$So $N^{0.01} > 100000$Giving $N > 100000^{100}$. Still impractical. The series $\sum \frac{1}{n(\ln n)^p}$ converges very slowly for $p$ near $1$.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.