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)

the 2nd one is just (3/2)^(k/2) so it of course diverges.

so, what do you get for arctan?

To determine the convergence of series, we can use the ratio test or the root test. Let's apply these tests to the given series:

1. arctan(e^-n):
Applying the ratio test, we consider the ratio of consecutive terms:
r = lim(n→∞) | arctan(e^-(n+1)) / arctan(e^-n) |
To simplify, notice that arctan is a monotonically increasing function, so we don't need to consider the absolute value.
Now, substitute in the terms:
r = lim(n→∞) arctan(e^-(n+1)) / arctan(e^-n)
Since the ratio involves arctan, which is a transcendental function, it becomes difficult to evaluate the limit analytically. In this case, we cannot definitively determine the convergence of the series using the ratio test.

2. ∑ (√3^k) / (2^k):
Applying the root test, we consider the limit of the k-th root of the absolute value of the terms:
r = lim(k→∞) ∛(| √3^k / 2^k |)
Simplifying further:
r = lim(k→∞) (∛(3^k) / 2^(k/2))
Notice that ∛(3^k) is equivalent to (∛3)^k, and 2^(k/2) is equivalent to (√2)^k.
r = lim(k→∞) [ (∛3 / √2)^k ]
Now, since (∛3 / √2) is a constant ratio, let's call it 'a'.
r = lim(k→∞) a^k
To determine the convergence of the series, we need to evaluate the value of 'a':
1. If |a| < 1, the series converges absolutely.
2. If |a| > 1, the series diverges.
In this case, (∛3 / √2) is approximately 1.078, which is greater than 1. Hence, the series ∑ (√3^k) / (2^k) diverges.

To summarize:
- The convergence of the series arctan(e^-n) cannot be determined using the ratio test.
- The series ∑ (√3^k) / (2^k) diverges.