Infinite Sequences and Series (BC Only)

Conduct Limit Comparison Test with convergent p-series, ineffective since comparison requires positive terms and understanding of limit behavior.

Use the Alternating Series Test after verifying that $\lim_{n \to \infty} a_n$ does not exist or equals infinity.

Employ Taylor Series expansion around $n=\infty$, which doesn't simplify analysis for simple power functions like $\sqrt{n}$.

Implement Partial Fraction Decomposition and test individual terms' limits, which is inappropriate for non-rational sequences.

It fails because its sequence's limit is not zero.

It passes because its sequence's limit is zero.

It fails because its partial sums decrease without bound.

It passes because its partial sums are bounded above.

It diverges based on comparison with geometric series whose ratio's absolute value exceeds one.

The terms go towards an indeterminate form suggesting further analysis or tests are necessary for conclusions on convergence/divergence.

It converges because each term gets smaller as n gets larger due to division by an exponential function in denominator.

No conclusion about convergence can be made since $\lim_{n\to\infty} a_n = \frac{5}{4}$, which is a nonzero constant.

It diverges because $\lim_{n \to \infty} b_n$ does not exist.

It converges because the terms alternate in sign and decrease in absolute value.

It converges since applying L'Hôpital's Rule shows that limit goes to zero.

Nothing can be concluded since $\lim_{n \to \infty} b_n = 0$.

Nothing conclusive since the limit of terms is zero.

The series oscillates indefinitely because the terms are alternating in sign.

The series converges because the terms approach zero.

The series diverges because the limit of terms is not one.

The series diverges.

There is no change in the convergence of the series.

The series converges conditionally.

The series converges absolutely.

It converges absolutely.

It still diverges.

It converges conditionally.

The test is inconclusive for this alteration.

$k=0$

$k=1$

$k=2$

$k=-1$

$-\infty$

$1$

$+\infty$

$0$

It converges because it's an alternating series with decreasing magnitude.

It diverges because $\lim_{n \to \infty} a_n \neq 0$.

It converges absolutely because $\sin\left( \frac{\pi}{n} \right) \to 0$ as $n \to \infty$.

Nothing can be concluded because $\lim_{n \to \infty} a_n = 0$ does not ensure convergence.