Infinite Sequence and Series

The Maclaurin series of a function f(x) is the Taylor series expanded about x = 0, given by: $$f(x) = f(0) + f'(0)\cdot x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots + \frac{f^{(n)}(0)}{n!}x^n + \cdots$$ --- Let f(x) = sin x. We compute each der...

The Maclaurin series is a special case of the Taylor series expanded about x = 0. It is given by: $$f(x) = f(0) + f'(0)\cdot x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots = \sum{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n$$ --- Since the derivative o...

- Part 1: Function $f(x) = \cos x$; find Maclaurin series and prove it represents $\cos x$ for all $x$. - Part 2: Define IVP; solve $y' + 2y = 3$, with initial condition $y(0) = 1$. - Part 3: Find the volume of a sphere of radius $r$. All required data is p...

(a) Maclaurin Series for $e^x$ and Proof The Maclaurin series of a function is: $$f(x) = \sum{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}\,x^n$$ For $f(x) = e^x$, all derivatives equal $e^x$: $$f^{(n)}(x) = e^x \implies f^{(n)}(0) = 1 \quad \text{for all } n$$ Ther...

Given: - Sequence: $an = (-1)^n$ No other numeric data required. This is a proof-based analysis problem. --- $$a1 = -1,\quad a2 = +1,\quad a3 = -1,\quad a4 = +1,\quad \dots$$ The sequence oscillates permanently between $+1$ and $-1$. A sequence $\{an\}$ con...

Given data: - Sequence general term: $an = \dfrac{n}{\sqrt{5+n}}$ - Task: define a sequence; determine convergence. All required data present. --- A sequence is a function whose domain is the set of natural numbers $\mathbb{N}$. It is an ordered list of ter...

- General term: $an = \dfrac{n}{\sqrt{5+n}}$, $n \in \mathbb{N}$ - Required: definition of a sequence; determine convergence of $\{an\}$ --- A sequence is a function whose domain is the set of natural numbers $\mathbb{N}$. It is an ordered list of numbers a...

- Series: $\displaystyle\sum{n=1}^{\infty} an$ where $an = \dfrac{n^2}{5n^2+4}$ - Task: determine convergence or divergence. Method: nth-Term Divergence Test. If $\displaystyle\lim{n\to\infty} an \neq 0$ (or does not exist), then $\sum an$ diverges. Compute...

Series to test: $$\sum{n=0}^{\infty} \frac{1}{1+n^2}$$ Term: $an = \dfrac{1}{1+n^2}$, starting at $n = 0$. No numerical parameters missing; this is an analytical convergence proof. --- Integral Test: If $f$ is continuous, positive, and decreasing on $[1,\in...

- Series: $\displaystyle \sum{n=1}^{\infty} an$ where $an = \dfrac{n^2}{5n^2 + 4}$ Test to apply: Divergence Test (nth Term Test) Theorem: If $\lim{n \to \infty} an \neq 0$ (or does not exist), then $\sum an$ diverges. Compute the limit of the general term:...

(a) Convergence of $\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ - Series: $\displaystyle\sum{n=1}^{\infty} \frac{(x-3)^n}{x}$ Since $x$ is independent of the summation index $n$, factor out $\frac{1}{x}$: $$\sum{n=1}^{\infty} \frac{(x-3)^n}{x} = \frac{1}{x}\sum{n...

Series to test: $$\sum{n=1}^{\infty} un, \qquad un = \frac{n^n}{n!}$$ Because the general term involves a factorial $n!$, D'Alembert's Ratio Test is most convenient. $$un = \frac{n^n}{n!}, \qquad u{n+1} = \frac{(n+1)^{n+1}}{(n+1)!}$$ Form the ratio: $$\frac...