Infinite Sequences and Series (BC Only)

Not sure where it moves since radial negative values aren't allowed.

It moves across the polar axis into the opposite quadrant.

The point gets closer to the fixed pole but stays in the same quadrant.

There is no change in position since the angle doesn't change.

It violates both Leibniz’s test for convergence and established bounds on accuracy based on following terms’ magnitudes indicating divergence or misapplication of tests/bounds.

Incorrect application of Leibniz’s test yet observes bounds rules thus implying potentially slower rate of convergence instead.

There is no violation or implication as long as subsequent terms continue decreasing which guarantees eventual conformance to established thresholds regardless of initial partial sums' exceedance.

Correct application Of Leibniz’s test indicating accelerated rate of convergence beyond expected parameters set By usual bounding principles.

$\frac{3}{64}$

$\frac{3}{81}$

$\frac{3}{9}$

$\frac{6}{65}$

$\theta$ can be any real number

$0 \leq \theta < 2\pi$

$0 < \theta \leq 360$

$-\pi < \theta \leq \pi$

Higher error bound indicates higher accuracy.

Higher error bound indicates lower accuracy.

Lower error bound indicates lower accuracy.

Lower error bound indicates higher accuracy.

More than or equal to a fraction

More than a fraction

Less than a fraction

Less than or equal to a fraction

The series converges faster due to the natural logarithm's properties.

Convergence may be affected since $\ln\left(\frac{n+1}{n}\right)$ doesn't necessarily decrease monotonically for all n.

Convergence isn't affected as long as all terms remain positive.

The series diverges because logarithmic functions have slower decay than polynomials or exponentials.

Greater than $\frac{\ln(5)}{25}$ but less than $\frac{\ln(6)}{36}$

Exactly $\frac{\ln(4)}{16}$

Less than or equal to $\frac{\ln(5)}{25}$

Equal to zero since it converges absolutely.

Less than or equal to $\frac{1}{36}$

Less than or equal to $\frac{1}{25}$

Less than or equal to $\frac{1}{49}$

Less than or equal to $\frac{1}{16}$

The error cannot be estimated without knowing exact values of subsequent terms.

The error will be greater than or equal to twice the absolute value of the first omitted term.

The error will be less than or equal to the absolute value of the first omitted term.

The error will oscillate between zero and twice the absolute value of any term.