Limit of a series exists but series diverges?

Question

In finding the sum of the series $$\sum_{n=1}^{\infty} \frac{4}{(4n-3)(4n+1)}$$ I get to $$1-\frac{1}{4n+1}$$

and I just take $$\lim_{n\to\infty} a_{n} = 1$$

which tells me that the sum of the series as n goes to infinity is 1. However, there's a theorem in the book I'm using (Thomas Calculus) that states that $\sum_{n=1}^{\infty} a_{n}$ diverges if $\lim_{n\to\infty} a_{n}$ fails to exist or is different from zero. T

So this would lead me to believe that the sum of the series is 1 but that it diverges. This doesn't make sense to me. Any help would be greatly appreciated!

EDIT: I see what my problem was now. I was writing out the first few terms in order to do telescoping to get the sum. However, the limit that I was taking was of the sum, NOT of the nth term. The nth term should go to zero. This makes sense. Thank you all for clarifying

Answer 1

Your theorem involves the limit of the terms, not the limit of the sum. Here $\lim_{n \to \infty} \frac 1{(4n-3)(4n+1)}=0$ so you have no problem. This does not guarantee that the sum converges, but it allows it.

Answer 2

The series converges. I think that the theorem that you have in mind is about the limit $\lim_{n\to\infty}\frac1{(4n-3)(4n+1)}$, which is $0$. Therefore, there is no contradiction.

Answer 3

Note that we represent $a_n$ as $n$-th term of the sequence. So, the correct equality is $$\lim_{n \to \infty} a_n=\lim_{n \to \infty} \frac 4{(4n-1)(4n+3)}=0$$