Infinite Sequences and Series (BC Only)

Divergent because limes superior equals −∞

It condenses by justifying means theory

Divergent by Theorem because limes superior equals ∞

It convegences because limes inferior equals ∞

$x = ae^{bt}$, $y = ce^{dt}$

$x = at + b$, $y = ct + d$

$x = a(\sin t + b)$, $y = c(\cos t + d)$

$x = a\tan(bt)$, $y = c\tan(dt)$

It approaches infinity as $n$ approaches infinity for some fixed $\alpha > 0$.

It oscillates between bounds above and below one as $n$ approaches infinity regardless of $\alpha$’s value.

It equals zero as $n$ approaches infinity for all values of $\alpha$.

It remains constant at any value other than one as $n$ approaches infinity for some fixed $\alpha > 0$.

$k > e$

$1 < k \leq e$

$k = e$

$0 < k \leq 1$

It diverges because the ratio exceeds $-3$ but is less than $0$ when applying limits as n approaches infinity.

It alternately converges and diverges due to periodicity in ratios across terms.

It converges because the ratio is less than $1$.

No definitive conclusion can be drawn; another convergence test may be necessary for a conclusive answer.

$(x'(t), y'(t))$

$(t \cdot x(t), t \cdot y(t))$

$\frac{dx}{dt}, \frac{dy}{dt}$

$\int x(t) dt, \int y(t) dt$

$k > 0$

$1 < k \leq e$

$k > e$

$0 < k \leq 1$

$L = \lim_{n \to \infty} \frac{(n+1)!/5^{n+1}}{n!/5^n}$

$L = \lim_{n \to \infty} \frac{n!/5^n}{(n-1)!/5^{n-1}}$

$L = 0$

$L = \lim_{n \to \infty} n!\cdot5^{-n}$

The series converges only at x = -3 and diverges otherwise.

The series converges for $x < 7$ and diverges for $x > 7$.

The series converges for all real numbers except x = -3 due to asymptotic behavior near that point.

The series converges when absolute value of $\frac{x-3}{4}$ is less than or equal to one-half ($\left| \frac{x-3}{4} \right| \leq \frac{1}{2}$).

$\int_{a}^{b}\left(\frac{y}{x}\right) dt$

$\int_{a}^{b}\sqrt{(x')^2 + (y')^2} dxdy$

$\int_{a}^{b}(x)dy = \int_{a}^{b}(x)\frac{dy}{dt} dt$

$\int_{a}^{b}(y)dx = \int_{a}^{b}(y)\frac{dx}{dt} dt$