RenerLi16

Problem: Luke studies $\sum_{n=1}^{\infty} \frac{(x-1)^n}{n}$: find the interval of convergence.

Solution: First, use the Ratio Test to find the interior of the interval: $$L = \lim_{n \to \infty} \left| \frac{\frac{(x-1)^{n+1}}{n+1}}{\frac{(x-1)^n}{n}} \right| < 1$$ $$L = \lim_{n \to \infty} \left| \frac{(x-1)^{n+1}}{n+1} \cdot \frac{n}{(x-1)^n} \right| = \lim_{n \to \infty} \left| (x-1) \frac{n}{n+1} \right|$$ Since $\lim_{n \to \infty} \frac{n}{n+1} = 1$, we get: $$|x-1| < 1$$ This yields $-1 < x-1 < 1 \implies 0 < x < 2$.

RenerLi16

Problem: Palpatine creates $\sum_{n=1}^{\infty} \frac{(x-2)^n}{5^n}$: find the radius of convergence.

Solution: Use the Ratio Test to find where the series converges: $$L = \lim_{n \to \infty} \left| \frac{\frac{(x-2)^{n+1}}{5^{n+1}}}{\frac{(x-2)^n}{5^n}} \right| < 1$$ $$L = \lim_{n \to \infty} \left| \frac{(x-2)^{n+1}}{5^{n+1}} \cdot \frac{5^n}{(x-2)^n} \right| < 1$$ $$L = \lim_{n \to \infty} \left| \frac{x-2}{5} \right| < 1$$

Since the limit is independent of $n$: $$\left| \frac{x-2}{5} \right| < 1 \implies |x-2| < 5$$

RenerLi16

Problem: Ahsoka studies $\sum_{n=1}^{\infty} (-1)^{n+1}\frac{1}{n^2}$: classify convergence.

Solution: To classify convergence, we must check if the series converges absolutely or conditionally. First, check the absolute value of the series: $$\sum_{n=1}^{\infty} \left| (-1)^{n+1}\frac{1}{n^2} \right| = \sum_{n=1}^{\infty} \frac{1}{n^2}$$

This is a p-series with $p=2$. Since $2 > 1$, the absolute series converges. Because the absolute series converges, the original alternating series is classified as absolutely convergent.

RenerLi16

Problem: Kylo Ren's rage is $\sum_{n=1}^{\infty} \frac{3^n}{n!}$: use the Ratio Test to determine convergence.

Solution: The Ratio Test evaluates $\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$. If $L < 1$, it converges absolutely; if $L > 1$, it diverges; if $L = 1$, the test is inconclusive.

Set up the limit: $$L = \lim_{n \to \infty} \left| \frac{\frac{3^{n+1}}{(n+1)!}}{\frac{3^n}{n!}} \right|$$ $$L = \lim_{n \to \infty} \left| \frac{3^{n+1}}{(n+1)!} \cdot \frac{n!}{3^n} \right|$$

RenerLi16

Problem: Jar Jar says $\sum_{n=1}^{\infty} \frac{1}{n}$ converges because terms shrink: is he correct?

Solution: Jar Jar is using flawed logic based on the $n$-th Term Test for Divergence. The condition $\lim_{n \to \infty} a_n = 0$ is necessary for convergence, but not sufficient. The series $\sum_{n=1}^{\infty} \frac{1}{n}$ is the harmonic series, which is a p-series with $p=1$. By the p-series test (or integral test), it diverges even though its terms shrink to zero.

Answer: No, he is incorrect. It is the harmonic series, which diverges.

RenerLi16

Problem: Yoda studies $\sum_{n=1}^{\infty} \frac{1}{n^3}$: determine whether it converges or diverges.

Solution: This is a p-series of the form $\sum_{n=1}^{\infty} \frac{1}{n^p}$. Here, $p = 3$. By the p-series test, the series converges if $p > 1$ and diverges if $p \leq 1$. Since $3 > 1$, the series converges.

Answer: Converges

RenerLi16

Problem: The Empire's clone count is $100 + 50 + 25 + 12.5 + \cdots$: find the infinite sum.

Solution: This is an infinite geometric series. The formula for the sum is $S = \frac{a}{1 - r}$. The first term $a = 100$. The common ratio $r = \frac{50}{100} = \frac{1}{2}$.

Since $|r| < 1$, the series converges. $$S = \frac{100}{1 - 1/2} = \frac{100}{1/2} = 200$$

RenerLi16

Problem: BB-8 approximates $\ln 2$ with $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots$: find max error after 4 terms.

Solution: By the Alternating Series Estimation Theorem, the maximum error $|R_n|$ when using $n$ terms of a convergent alternating series is less than or equal to the absolute value of the first omitted term ($|a_{n+1}|$).

The series is $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$. We are using 4 terms, so the first omitted term is the 5th term ($a_5$). $$a_5 = \frac{1}{5}$$

Answer: $\frac{1}{5}$ (or $0.2$)

RenerLi16

Problem: C-3PO estimates $e^{0.5}$ using a third-degree Maclaurin polynomial: find the Lagrange error bound.

Solution: The Lagrange error bound for $P_n(x)$ is given by $|R_n(x)| \leq \frac{M}{(n+1)!} |x - c|^{n+1}$, where $M$ is the maximum value of $|f^{(n+1)}(z)|$ on the interval between $c$ and $x$. Here, $f(x) = e^x$, $n=3$, $c=0$, and $x=0.5$.

1. Find the $(n+1)$th derivative: $f^{(4)}(x) = e^x$.

2. Find $M$ on the interval $[0, 0.5]$:

RenerLi16

Problem: Finn approximates $\ln x$ centered at $x=1$: find the second-degree Taylor polynomial.

Solution: The Taylor polynomial centered at $x=c$ is $P_n(x) = f(c) + f'(c)(x-c) + \frac{f''(c)}{2!}(x-c)^2 + \dots$ Here, $c=1$ and $f(x) = \ln x$.

Find the derivatives at $x=1$: $f(x) = \ln x \implies f(1) = 0$ $f'(x) = \frac{1}{x} = x^{-1} \implies f'(1) = 1$ $f''(x) = -1x^{-2} = -\frac{1}{x^2} \implies f''(1) = -1$