Question

determine whether the series converges or diverges. show all your work clearly and state the test used.
(a) $sum_{n=1}^{infty} \frac{n^3}{2n^4 - 1}$
(b) $sum_{n=1}^{infty} (-1)^{n + 1} \frac{n}{e^{-n}}$
be sure to show all your work clearly and enter your answer.
(a) _________________________
(b) _________________________

Explanation:

Part (a)

Step 1: Choose Test (Limit Comparison)

We use the Limit Comparison Test. Compare $\sum_{n = 1}^{\infty}\frac{n^{3}}{2n^{4}-1}$ with $\sum_{n = 1}^{\infty}\frac{1}{n}$ (a p - series with $p = 1$, which diverges).

Step 2: Compute Limit

Calculate $\lim_{n
ightarrow\infty}\frac{\frac{n^{3}}{2n^{4}-1}}{\frac{1}{n}}=\lim_{n
ightarrow\infty}\frac{n^{4}}{2n^{4}-1}$. Divide numerator and denominator by $n^{4}$: $\lim_{n
ightarrow\infty}\frac{1}{2-\frac{1}{n^{4}}}=\frac{1}{2 - 0}=\frac{1}{2}$.

Step 3: Apply Test

Since the limit is a positive finite number and $\sum\frac{1}{n}$ diverges, by the Limit Comparison Test, $\sum_{n = 1}^{\infty}\frac{n^{3}}{2n^{4}-1}$ diverges.

Step 1: Simplify the Series

First, simplify $a_{n}=\frac{n}{e^{-n}}=n e^{n}$.

Step 2: Check nth - Term Test

Compute $\lim_{n
ightarrow\infty}a_{n}=\lim_{n
ightarrow\infty}n e^{n}$. As $n
ightarrow\infty$, $e^{n}$ grows exponentially and $n$ grows polynomially. So, $\lim_{n
ightarrow\infty}n e^{n}=\infty
eq0$.

Step 3: Apply nth - Term Test

By the nth - Term Test for Divergence, if $\lim_{n
ightarrow\infty}a_{n}
eq0$, then $\sum a_{n}$ diverges. So, $\sum_{n = 1}^{\infty}(-1)^{n + 1}n e^{n}$ diverges.

Answer:

(a):
The series $\sum_{n = 1}^{\infty}\frac{n^{3}}{2n^{4}-1}$ diverges (using the Limit Comparison Test with $\sum\frac{1}{n}$).

Part (b)