Calculus 2

Consider the series that starts from one goes to infinity: $(-1)^n \cdot \frac{1}{n}$. Will the alternating harmonic series converge or diverge?

Consider the series which goes from 1 to infinity: $(-1)^{n+1} \cdot \frac{5n + 3}{2n - 7}$. Will this series converge or diverge?

Consider the series: $\frac{(-1)^{n+1}}{5^n}$. Will the series converge or diverge?

Consider the series $(-1)^n \cdot \frac{n}{\ln(n)}$. Will the series converge or diverge?

Let's say the series is $(-1)^{n+1} \cdot \frac{\ln(n)}{n}$. Will it converge or diverge?

Consider the series $\frac{\cos(n \pi)}{n}$. Can we apply the alternating series test to it?

Take the series $(-1)^{n-1} B_n$, where $B_n$ is positive. Determine if the series is convergent or divergent based on if $B_{n+1} \leq B_n$ and $\lim_{{n \to \infty}} B_n = 0$.

Apply the alternating series test to different series to determine convergence or divergence: $(-1)^n \cdot \frac{3n-1}{2n+1}$ and $(-1)^{n+1} \cdot \frac{n^2}{n^3+4}$.

Use the ratio test to determine the convergence of the series $\frac{n^n}{n!}$.

Given an alternating series $\sum (-1)^{n-1} \frac{1}{n}$, determine if the series converges using the Alternating Series Test.

Determine if the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{4n-1}$ converges using the alternating series test.

Determine if the series $\sum_{n=1}^{\infty} \frac{n+1}{3n+2}$ converges or diverges, and justify your answer.

Determine if the series $\sum_{n=1}^{\infty} \frac{n}{n^2+1}$ converges using the alternating series test.

Determine if the series $\sum_{n=1}^{\infty} \frac{1}{\sqrt{2n+1}}$ converges using the alternating series test.

Approximate the sum of the series $\frac{(-1)^{n+1}}{n^2}$ correct to two decimal places.

Approximate the sum of the series $(-1)^{n+1}/n^3$ correct to three decimal places.

Determine the convergence or the divergence of the series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{4 + 3^n}$.

Determine the convergence or the divergence of the series $\sum_{n=1}^{\infty} \frac{1}{n^3 + 5}$.

Use the direct comparison test to determine if the series                                                             u000f n=1 }                                                               $n=1}^{        {   {