Very simple sequence/series question

Question

I have been stumped trying to figure out this question on my homework for some time now. I am required to find a convergent sequence a(n) where its counterpart f(x) that shares the same formula is divergent. I'm not sure if this is some sort of a typo or if I just can't find a solution:

Give a sequence that satisfies the following:

1. a_n = f(x) (that is f(x) and a_n have the same formula)
2. lim_n⇒∞ a_n converges
3. lim_n⇒∞ f(x) diverges

Answer 1

There are several different functions that could fit this, but I understand where your confusion is coming from; it seems at first like an impossible task. The trick, though, is to remember that convergence for a sequences and convergence for a function are different, and that all we know for sure is that the function has to pass through the every point in the series; all the other points are unknown.

The second important point is that divergent does not necessarily mean that it goes off to infinity; it would still be divergent as long as it has the property that it doesn't converge to a value. For instance, consider the function f(x)=1/x+cos(2πx). Clearly, for any integer value of x, this will just evaluate to 1/x; therefore, our sequence a_n = 1/n+cos(2πn) = 1/n, which converges to 0 as n goes to infinity. However, you can see that the general function f ever converges on a single value, instead oscillating between -1 and 1+ε. If you want, you can also define a function that will have an arbitrarily large value for certain larger values of x that still maintains the desired property, but I will leave that as an exercise to the reader.

Answer 2

How about this: a_n = cos(2πn) and its divergent counterpart f(x) = cos(2πx).

The sequence "converges", in the sense that it has a limit as n→∞, namely 1, but because f(x) is defined over the reals, the cosine function f(x) retains its periodicity, cycling through the y-values between -1 and 1 inclusive with a period of 1, and therefore is divergent (which is to say it is bounded, but the limit does not exist).