determine whether the series converges of diverges

the sum from k=1 to infinity of

cos(e^-k)

I'm not sure where to start..

As k becomes very large, e^-k approaches zero, and every cos(e^-k) term approaches 1.

You end up with an infinite series of terms all approaching +1, which cannot converge

To determine whether the series converges or diverges, you can use the convergence tests. One useful test is the Comparison Test, which compares the given series with a known convergent or divergent series.

In this case, let's use the Comparison Test and compare the given series with a simpler series that we can evaluate easily.

Consider the series:

sum from k=1 to infinity of |cos(e^(-k))|

Since the cosine function is bounded between -1 and 1, we can remove the absolute value signs without changing the convergence behavior. Therefore, we have the inequality:

|cos(e^(-k))| ≤ 1

Now, let's consider the series:

sum from k=1 to infinity of 1

This is a simple geometric series with a common ratio of 1, and it is well-known that this series diverges.

Now, using the Comparison Test, we compare the original series:

sum from k=1 to infinity of cos(e^(-k))

to the series:

sum from k=1 to infinity of 1

If the original series is smaller or equal to the second series, then it must converge. However, if the original series is greater than the second series, then it must diverge.

By comparing the terms, we have:

cos(e^(-k)) ≤ 1

Since cos(e^(-k)) is always between -1 and 1, it is less than or equal to 1. Therefore, the original series is smaller or equal to the series of 1.

As the series of 1 diverges, we can conclude that the original series, sum from k=1 to infinity of cos(e^(-k)), also diverges.

So, the series diverges.