Infinite Sequences and Series (BC Only)

When $p>1$.

Only when $p=1$.

When $0<p<1$.

For all positive values of $p$.

$\frac{1}{4}$

$\frac{1}{2}$

2

The series is not a p-series

$\sum_{n=0}^{\infty} \frac{1}{n^2}$

$\sum_{n=0}^{\infty} \frac{1}{n}$

$\sum_{n=1}^{\infty} \frac{1}{n}$

$\sum_{n=1}^{\infty} \frac{1}{n^2}$

The series may or may not converge depending on whether $(p-0.5) > 1$.

The convergence of the series remains unaffected by this change in $p$.

The series will always converge since it initially converged for $p > 1$.

The series will always diverge regardless of the value of $p$.

Harmonic series, $\sum_{n=1}^{\infty} \frac{1}{n}$.

P-series where $p=\frac{3}{2}$ .

Geometric series where ratio r satisfies $|r|<1$.

Alternating harmonic series, $\sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n}$.

A p-series is a type of harmonic series

A harmonic series is a p-series with p-value of 1

There is no relation between a p-series and a harmonic series

All p-series are also harmonic series, but not all harmonic series are p-series

$-1 < p < 0$

$p > 1$

$p = 0$

$0 < p < 1$

The transformed sequence still maintains conditional convergence similar to original alternating sign property's influence.

It causes divergence as natural logarithms grow slower than linear functions which alters summability negatively.

It has no significant impact as both sequences tend towards infinity and have diminishing terms contributing less over time.

It creates absolute convergence due to slow growth rate of natural logarithms compared to harmonic terms.

It diverges because its nth term does not approach zero.

It converges because terms decrease in absolute value.

It converges by comparison to a known convergent series.

Convergence cannot be determined without further analysis.

The terms in the series approach zero as $n$ approaches infinity.

The sequence of partial sums diverges to infinity.

The sequence of partial sums oscillates between two bounds.

The sequence of partial sums approaches a finite limit.