a. Consider the following limit as a fact:

lim n-> infinity ((n!)^1/n)/n = 1/e

Use this limit to study the convergence of this series using the root test.
Sum of infinity and n=1 of ((3^n)n!)/n^n

b. Use the ratio or the root test to determine the convergence of the series:
Sum of infinity and n=1 of arctan(e^-n), Sum of infinity and k=1 of sqroot(3^k)/(2^k)

a. To study the convergence of the series using the root test, we need to take the nth root of each term and look at the limit as n approaches infinity.

Consider the term ((3^n)n!)/n^n. Taking the nth root, we get:

((3^n)n!)/n^n)^(1/n)
= [((3^n)n!)^(1/n)] / (n^n)^(1/n)
= [(3^n)^(1/n) * (n!)^(1/n)] / n

Now, applying the limit we were given:
lim n-> infinity (n!)^(1/n) / n = 1/e

We can rewrite the expression as:
lim n-> infinity ((3^n)^(1/n) * (n!)^(1/n)) / n
= lim n-> infinity (3^n)^(1/n) * lim n-> infinity (n!)^(1/n) * lim n-> infinity 1/n

The first two limits are equal to 1/e and the third limit is 0. Therefore, we have:

(1/e) * 1 * 0 = 0

Since the limit is less than 1, the series converges by the root test.

b. Let's consider each series separately using the ratio and root tests:

For the series: Sum of infinity and n=1 of arctan(e^-n)

Ratio Test:
Taking the ratio of consecutive terms, we have:
| (arctan(e^-(n+1))) / (arctan(e^-n)) |
= | (e^-(n+1)) / e^-n |
= | e^-(n+1) / e^-n |
= | e^-(n+1 + (-n)) |
= | e^(-1) |
= e^(-1)

Since the limit of the ratio is less than 1, the series converges by the ratio test.

Root Test:
Taking the nth root of the terms, we have:
(arctan(e^-n))^(1/n) = (e^(-n))^(1/n) = e^(-1)

Since the limit of the nth root is less than 1, the series converges by the root test.

For the series: Sum of infinity and k=1 of sqroot(3^k)/(2^k)

Ratio Test:
Taking the ratio of consecutive terms, we have:
| (sqroot(3^(k+1))/(2^(k+1))) / (sqroot(3^k)/(2^k)) |
= | (sqroot(3^(k+1)) / 2^(k+1)) * (2^k / sqroot(3^k)) |
= | (sqroot(3^(k+1) * 2^k)) / (2^(k+1) * sqroot(3^k)) |
= | sqroot((2 * 3^k) / (2 * 3^k)) |
= 1

Since the limit of the ratio is equal to 1, the ratio test is inconclusive.

Root Test:
Taking the nth root of the terms, we have:
(sqroot(3^k) / 2^k)^(1/k) = (3^k)^(1/(2k)) / (2^k)^(1/k) = (3^k)^(1/(2k)) / 2

Letting n tend to infinity (which means k also tends to infinity), we see that the limit is equal to 1/2. Since the limit is less than 1, the series converges by the root test.

In conclusion, both series converge: the series of arctan(e^-n) converges by both the ratio and root tests, while the series of sqroot(3^k)/(2^k) converges by the root test but the ratio test is inconclusive.

a. To study the convergence of the series using the root test, we need to determine the limit of the absolute value of the nth root of the terms in the series as n approaches infinity.

Let's apply the root test to the series:
Sum of infinity and n=1 of ((3^n)n!)/n^n

First, we take the absolute value of each term:
|((3^n)n!)/n^n|

Next, we calculate the limit of the absolute value of the nth root of the terms as n approaches infinity:
lim n-> infinity [(|((3^n)n!)/n^n|)^(1/n)]

We can rewrite the numerator as (n!)^1/n and the denominator as n.

Now, we can rewrite the expression as:
lim n-> infinity [(n!)^1/n/n]^(1/n)

Using the given limit as a fact:
lim n-> infinity [(n!)^1/n/n]^(1/n) = 1/e^(1/n)

Since 1/n approaches 0 as n approaches infinity, we can evaluate the limit further:
lim n-> infinity 1/e^(1/n) = 1/e^0 = 1/e

Therefore, the limit of the absolute value of the nth root of the terms in the series is 1/e.

According to the root test, if the limit is less than 1, the series converges. If it is greater than 1 or infinite, the series diverges. Since the limit is 1/e which is less than 1, the series converges.

b. Let's consider each series separately and apply the ratio and root tests to determine their convergence.

1. Series: Sum of infinity and n=1 of arctan(e^-n)
Using the ratio test:
We calculate the absolute value of the ratio of consecutive terms:
|(arctan(e^-(n+1))) / (arctan(e^-n))|

Simplifying the expression, we get:
|arctan(e^-(n+1)) / arctan(e^-n)|

The ratio test states that if the limit of the absolute value of this ratio as n approaches infinity is less than 1, the series converges.

2. Series: Sum of infinity and k=1 of sqrt(3^k) / (2^k)
Using the ratio test:
We calculate the absolute value of the ratio of consecutive terms:
|sqrt(3^(k+1))/(2^(k+1)) / sqrt(3^k)/(2^k)|

Simplifying the expression, we get:
|sqrt(3^(k+1))/sqrt(3^k) * (2^k)/(2^(k+1))|

Notice that sqrt(3^(k+1))/sqrt(3^k) simplifies to sqrt(3).

The ratio test states that if the limit of the absolute value of this ratio as k approaches infinity is less than 1, the series converges.

By applying the ratio or root test to each series separately, you can determine the convergence of both.