Infinite Sequences and Series (BC Only)

No, because $\lim\limits_{n \to \infty} \frac{(-1)^n}{\ln(n)}$ does not exist.

Yes, because $\lim\limits_{n \to \infty} \frac{1}{\ln(n)} = 0$ and $\frac{1}{\ln(n)}$ is decreasing.

No, because $|\frac{(-1)^n}{\ln(n)}|$ increases as $n$ increases.

Yes, because $\lim\limits_{n \to \infty} (-1)^n = -1$, indicating convergence.

INCORRECT. The terms ${a_n}$ must increase without bound.

CORRECT. The terms ${a_n}$ must converge to zero while $\sum a_n$ diverges.

INCORRECT. The sequence ${a_n}$ should alternate signs independently from factor (-i)n.

INCORRECT. All terms in the sequence ${a_n}$ should be equal.

1 - 1/3 + 1/5 - 1/7 + ...

1/2 + 1/4 + 1/6 + 1/8 + ...

1 - 2 + 3 - 4 + ...

1/2 + 2/3 + 3/4 + 4/5 + ...

The terms increase without bound as n increases.

The nth term does not approach zero as n approaches infinity.

Every term is greater than the sum of all subsequent terms.

The terms decrease monotonically to zero.

$\sum (-1)^{n}(0.5)^{n}$

$\sum (-1)^{n}(\ln(n))$

$\sum (-1)^{n}(0.5)^{n}$

$\sum (-1)^{n}(e^n)$

The series $(-1)^n$ and the terms decrease and alternate signs.

The series $S(-1)^n$, where the terms alternate signs but don't necessarily decrease.

The series $E(-1)^n$, where $\lim_{n \to \infty} a_n$ is not zero.

The series $(-1)^n$ and the terms decrease monotonically.

1 - 2 + 3 - 4 + ...

1 + 2 + 3 + 4 + ...

1 - 1/2 + 1/3 - 1/4 + ...

1 + 1/2 + 1/4 + 1/8 + ...

Compare derivatives of numerator and denominator functions within a corresponding sequence form to infer about convergence using related rates.

Convert the summation into an integrable function and then use L'Hôpital's Rule on resulting indeterminate forms after integration.

Directly apply L'Hôpital's Rule on sequence terms treating them as functions over continuous variables until reaching a solvable limit form.

Utilize L'Hôpital's Rule iteratively on each term of the series expansion until exhibiting clear patterns that suggest behavior at infinity.

When terms approach zero and decrease monotonically.

When every term is larger than the previous term.

When all its terms are greater than one.

If it can be expressed as a geometric series.

The series diverges because it fails to meet these conditions beyond this specific term.

The series definitely converges because it meets conditions at a particular term.

It's indeterminate until we test convergence using another method such as Ratio or Root Tests.

Additional analysis is needed since not all terms have been tested yet.