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Suppose that $X_n = n^2$ with probability $\frac{1}{n}$ and $0$ otherwise with probability $1-\frac{1}{n}$, I would like to show that $X_n$ converges in probability to $0$. Hence, it suffices to show that for any given $\epsilon >0$, $$ \lim_{n\to \infty} P(|X_n-0|>\epsilon) =0 $$

To do this, I obviously will find $P(|X_n|> \epsilon)$. The way I do this is to recognize that $P(|X_n|> \epsilon) = 1-P(|X_n|< \epsilon) = \frac{1}{n}$.

HOWEVER: When I try to directly find $P(|X_n|> \epsilon)$, there is always the chance that I can set $\epsilon$ to be greater than whatever $X_n$ I am currently at in the sequence, in other words, if I made $\epsilon$ to be $1000$, and $n=4$, what would the probability be in this case?

It seems it is no longer $\frac{1}{n}$ BUT $0$. It seems like we cannot compute $P(|X_n|> \epsilon)$ for ALL $\epsilon>0$. What am I missing here?

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It looks to me like you've fallen prey to a common problem - you're trying to solve the problem before you've completely stated it. You say you're trying to show that "$P(|X_n - 0| > \epsilon)$" for all $\epsilon > 0$; that's like saying that you're trying to show "$4$".

What I think you meant to say is that you're trying to show that for every $\epsilon > 0$, $\lim_{n \to \infty}P(|X_n - 0| > \epsilon) = 0$. The important thing to notice is the order of the quantifiers: for every $\epsilon$ there is a large enough $n$ so that (etc). That means that you do not have the liberty to choose $\epsilon$ after $n$ has been selected; you must select $\epsilon$ first, and then a large enough $n$ will be supplied. So, in your example of $\epsilon = 1000$, $n = 4$ may not work - but $n = 1000$ does, and that's good enough for limits. (Also, as noted in Lee's answer, for such wild disparities of $n$ and $\epsilon$ we just get a free $0$, which supports the limit anyway, so there was never a problem at all.)

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  • $\begingroup$ Nice answer, +1. $\endgroup$ – YoTengoUnLCD Dec 13 '16 at 3:48
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What you want to show is

$$A_n\equiv P(|X_n-0|>\epsilon) \quad \lim_{n \to \infty}A_n = 0 \qquad \text{for any $\epsilon >0$}$$

You asked

if I made $\epsilon$ to be 1000, and $n=4$, what would the probability be in this case? ... It seems it is NO LONGER $1/n$ BUT $0$

Yeah, so in such big-$\epsilon$ cases, you already have $A_n$ identically zero starting at some finite $n$ (the rest of the sequence of $A_n$ are all exactly zeros), not just $\lim A_n \to 0$. In such cases the probability $A_n$ is trivially computed. (why do you think we cannot compute it?)

Usually when people talk about calculation to show the convergence, it is the vanishingly small $\epsilon \to 0$ cases that need some work.

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  • $\begingroup$ oh I misread the question statement and my answer does not apply to the situation at all: for a given $\epsilon$ (large or not), larger $n$ will have a tiny mass of $1/n$ outside of the bound of $\epsilon$. Namely, $A_n = 1/n$ and NOT identically zero. What user A.G. said is correct, just the the R.H.S. should be $\leq 1/n$ $\endgroup$ – Lee David Chung Lin Oct 8 '17 at 22:57
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First have a look at the definition of convergence in probability: it is for any given $\epsilon>0$. Once $\epsilon$ is given then $P(|X_n|>\epsilon)$ is either $0$ or $1/n$, at any rate $$ 0<P(|X_n|>\epsilon)<\frac 1n $$ which is more than enough to show that it converges to $0$ as $n\to\infty$.

Btw, if you need to get back to the definition of convergence for a series, note that you will need a different $\epsilon$ (that you can call $\epsilon'$ for instance).

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