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So, we know that a geometric series is convergent if -1 < r < 1. Right? After deciding that our r fits in this range, we can use the formula to see what it actually converges to. So, we can find that our sequence converges to some number.

BUT...

How is it that we then say that if a series is convergent, the limit of its related sequence must be 0. (If the limit is NOT 0, it is divergent. If the limit IS 0, we still know nothing about convergence or divergence)

Q1. So, are we, now, saying that the limit of a series equaling a real number does not prove its convergence?

Q2. Did we ever say that it did, or am I getting sequences and series mixed up?

If I sound confused, it's because I am!

Update: I think I have pinpointed my confusion:

We can test the convergence of a series by checking to see if the limit of the sequence of partial sums converges.

But, with the "Test for Divergence", we test to see if the limit of the sequence = 0.

  1. If it does not equal 0, our series diverges.
  2. If it does, we know nothing.

So, is the key point here that for one test, we are testing the sequence of partial sums. And for another, we are just testing the terms in series, written as a sequence with commas instead of +'s?

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    $\begingroup$ It seems that you are confusing the sequence $a_n$ with the series $\sum_n a_n$. If the latter converges (i.e. it is equal to a real number!), then the first converges to $0$. Nothing but this. $\endgroup$ – Crostul Nov 16 '16 at 13:19
  • $\begingroup$ Okay, but don't we use the limit of the sequence to determine whether the series equals a real number or is convergent ...? In other words, how do we test the convergence of a series then? $\endgroup$ – LaSpana101 Nov 16 '16 at 13:29
  • $\begingroup$ The condition that $(a_n)$ tends to $0$ is necessary, but not sufficient. There are various methods to check the convergence of a series, for example the comparison test. $\endgroup$ – Peter Nov 16 '16 at 13:34
  • $\begingroup$ @Peter No. $a_n\to0$ and $a_n$ decreasing is sufficient.$$\sum_{n=2}^\infty\frac{(-1)^n}{\sqrt{n}+(-1)^n}$$ is a counterexample. $\endgroup$ – Julián Aguirre Nov 16 '16 at 14:48
  • $\begingroup$ @JuliánAguirre Correct, thanks for pointing it out. $\endgroup$ – Peter Nov 16 '16 at 14:52

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