Determine whether each of the series below?

Only b) is possibly conditional convergence, you need an x or some variable (k is the index NOT a variable) for conditional convergence (n is the variable in part b)

a)

Do you mean: atan(k) / k², because what you appear to have written is atan(1/k)

I'm not real sure how to do this one, I suspect that it converges, because the integral of arctan(x) involves a ln(....), sot his suggests that it's on the same order as 1/x ...actually maybe the squeeze theorem is how to do this:

|atan(x)| < π/2, for ALL x -->> therefore the following CONVERGENT series is greater than this one:

π/2/k² --> this series converges (because it's a p-series with exponent > 1, or < -1 depending on how you write it).

Since every term is less than the term of the convergent series, the original series must ALSO converge. If a series converges, and you look at a series that is clearly less than it, then the other series (the arctan in our case) either converges or diverges to -∞, but since our series is always > 0, we know it converges:

Squeeze theorem:

atan(k)/k² > 0 --> ∑0 --> converges

atan(k)/k² < π/2/k² -->∑1/k² --> converges

So you have bounded this series by two convergent series, therefore it also converges

b)

This is just a p-series, the n doesn't make a difference, it will converge no matter what n you choose...assuming n is an integer, if it's a real number, (-1)^n is difficult to do, but it will still converge.

c)

I assume you mean:

(5^k + k)/(k! + 3), if you actually mean what you wrote: 5^k + (k/k!) + 3, then this of course diverges

This will probably converge, I would again use squeeze theorem, because the + k and the + 3 aren't really important, so look @ following series:

(5^k) / (k!) --> check using ratio test:

(5^(k + 1) / (k^k)) / ((k + 1)! / k!) = 5 / k --> 0 as k goes to infinity, therefore this series converges

So now make the case that:

5^k + k > 5^k, not very helpful

On the other hand:

(k! + 3) > k! therefore (5^k)/(k! + 3) < (5^k)/(k!), therefore 5^k / (k! + 3) convereges

but we HAVE (5^k + k)/(k! + 3) = (5^k)/(k! + 3) + k / (k! + 3)

Well we've already shown that the 1st converges...in fact we use a dirty trick to not do it separately (which you should be able to show):

5^k + k < 5^k + 5^k = 2 * (5^k)

Well 2 * (5^k) / (k! + 3) certainly converges so our original converges.

Edit: I'm pretty sure you have to do limit comparison for all of these, which is the most difficult to do because it's hard to choose the correct, convergent or divergent series to compare with. For instance it's always easy to find a divergent series that's greater, and it's also easy to find a convergent series that is less than, but it's the others that are not so straight forward.

Here's b) worked out:

I don't know what you mean here?

Here's what you wrote:

{(-1)^n} + {1 * 3^(2n - 1) / k²} + 1

of course this diverges, when n is even, converges when n is odd

I think you meant to write:

(-1)^(n + 1) * 3^(2n - 1) / (k² + 1)

The top part, evaluates to some number no matter what k you choose:

A/(k² + 1)

This definitely converges because if you compare to:

1/k² (which converges, p-series)

well k² + 1 > k² therefore 1/(k² + 1) < 1/k², therefore this series is less than a convergent series, therefore it converges.

The constant just multiples the final value of the series, so the positive or negative can just be factored out of the sum and then you don't have to deal with it at all.

Edit2:

By the way, good luck, this is probably the hardest topic in Calc. III, I know I didn't understand it very well until I started tutoring in college.