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I haven't been able to find any sources that clearly and methodically state the approach for proving the convergence of a sequence, using the epsilon definition of convergence. At best, I have been able to find vague, unjustified demonstrations. However, this does nothing to help me learn. I want to be able to generalise this method across all convergence problems that I encounter.

I would like someone to state the steps and associated reasoning involved in proving that a sequence converges, using the epsilon definition of convergence. Please specify the reasoning behind each step of the methodology, to assist in justifying your calculations. I would like the 'why' and 'how' behind each step of such a proof.

I have the sequence $ \{a_n\}_{n=1}^{\infty}$, where $a_n = \dfrac{(-1)^{n+1}}{n}$, $L = 0$.

From what I have read, we want to prove that for any $\epsilon > 0$, there exists some $N > 0$, such that if $n > N$, $|a_n - L| < \epsilon$. However, as alluded to above, I do not fully appreciate or understand what this is saying.

Thank you.

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    $\begingroup$ I think there is a mistake while specifying $a_n$. Please check. $\endgroup$ Nov 5, 2016 at 9:47
  • $\begingroup$ @астонвіллаолофмэллбэрг Indeed, there was. Thank you $\endgroup$ Nov 5, 2016 at 9:49
  • $\begingroup$ The sum does not go to zero. It is only $a_n \to 0$, not $\sum a_n \to 0$, right? $\endgroup$ Nov 5, 2016 at 9:51
  • $\begingroup$ Convergence is more easily proven with the alternating series test. $\endgroup$
    – Sophie
    Nov 5, 2016 at 9:52
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    $\begingroup$ @ThePointer That's great. I instantly knew the answer was non-zero because I had seen the series before, it vaguely reminded me of $\ln$. Great then, but proving this is more difficult, it requires good knowledge of properties of $\ln$ , Taylor series etc. $\endgroup$ Nov 6, 2016 at 0:41

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To be able to generalize procedures across various epsilon-delta proofs, it is important to notice what are the stand out features of such proofs(tricks, conversions etc.)

In this case, suppose we want to show that $L=0$. Let us take $\epsilon>0$. We want to find an $N >0 $ such that if $n > N$ then $|a_n - L| < \epsilon$. In our case, $L=0$, so it changes to $|a_n| < \epsilon$.

Step 1: Look at $a_n$ carefully. It is $\dfrac{(-1)^{n+1}}{n}$. Now, we are required to find out about $|a_n|$, so let us compute $|a_n|$. This is a step by itself, because it gives a clear direction of attack: To attack this problem, we will calculate $|a_n|$ explicitly, and try to find $N$ satisfying the limit conditions explicitly. We normally do this because $a_n$ is not a very complicated quantity, so it is easy to work with.

Step 2:So what is $|a_n|$? It is $\left|\dfrac{(-1)^{n+1}}{n} \right|$. Since the modulus splits across the fraction, $$ \left|\dfrac{(-1)^{n+1}}{n} \right| = \dfrac{|(-1)^{n+1}|}{|n|} = \dfrac{1}{n} $$ There is no trick here. We just simply calculated $|a_n-L|$ directly, because in this problem it was easy to do so. The reason for this step is motivated by the previous step.

Step 3: Now, suppose we were given $\epsilon>0$ and were asked to find a large enough $N$ such that if $n > N$ then $|a_n| < \epsilon$. However, we have now calculated $|a_n|$, and it is $\frac 1n$. Hence, we are trying to find $n$ such that $\frac{1}{n} < \epsilon$. However: $$ \epsilon > \frac 1n \iff n > \frac{1}{\epsilon} $$ The above piece of insight is vital to us: we can find our $N$ explicitly.

Step 4: Let $N$ be the smallest integer greater than $\frac{1}{\epsilon}$. Then, note that: $$ n > N \implies n > \frac 1 \epsilon \implies \epsilon > \frac 1n $$

Step 5: Hence, this $N$ works for the given problem, so we can conclude by definition of limit that $a_n \to 0$.

What have we learnt from here?

1) Wherever possible and easy, calculate $|a_n -L |$ explicitly. It is the safest option for simple looking $a_n$.

2) In Step $3$, we actually worked backwards. We assumed that the $N$ which we wanted existed, and then we tracked back to actually find that $N$, right? Working backwards is a very big trick, because the $N$ that you want can be explicitly found via working backward, as has happened in this case.

Unfortunately, the problem that you have is not very illustrative, because it doesn't go through all the tricks and twists that one goes through while evaluating a normal tricky limit. You need to find a good example (and me), and then you will get a better grip on limits, because (certainly unknown to you, so please don't berate yourself) this problem was damp squib compared to some of the harder limits you have to come across. Hopefully, though, I have done justice to this problem. Please get back on doubts.

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  • $\begingroup$ Your explanation is excellent and exactly what I was looking for. Thank you so much for posting such an illuminative response. Hopefully I will encounter a more educational example in the future. $\endgroup$ Nov 5, 2016 at 10:36
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    $\begingroup$ @ThePointer You will have many more complicated examples: rational functions, trigonometric functions etc., and of course, L'Hopital. Would love to go through some of them with you. $\endgroup$ Nov 5, 2016 at 10:39

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