Prove the sequence has no limit

Question

I want to prove the sequence $a_n = (-1)^n + 1/n$ has no limit for $n\to +\infty$.

Here is my attempts.

By the definition, let's suppose $a_n$ has a limit, and call it $\ell$. Since for $n\to +\infty$ we have $(-1)^n + 1/n$ = \pm 1$ we intuitively know there is no limit. But let's suppose there is.

Let's say $\ell = 1$ and $n$ is even. Then

$$|(-1)^n + 1/n -\ell | < \epsilon \implies |1 + 1/n - 1| < \epsilon$$

This means $1/n < \epsilon$. For this case it seems like the sequence has limit though.

But if I choose $\ell = 1$ and $n$ odd, then I get

$$|-1 + 1/n - 1| < \epsilon \implies |1/n - 2| < \epsilon$$

Now here we can see that as $n\to +\infty$, $1/n - 2 \to -2$ and hence $|-2| = 2 < \epsilon$ means $\epsilon > 2$, which contradicts the limit definition "$\forall \epsilon$".

I don't know if this is correct or not, but I am wondering why in the first choice I get something that appears all good. Perhaps it's because the limit is indeed $\ell = 1$ when $n$ goes to infinity "in an even way"...?

But this would prove the limit exists.

If I do the same thing for $\ell = -1$ and $n$ odd, I too get the limit exists.

How to interpret the two existences? Maybe I shall invoke the theorem for which if the limit exists then it should be unique?

Thank you so much!

Answer 1

To complement John Hughes' (+1) answer and the comments, we can apply the reverse triangle inequality to arbitrary consecutive terms, deducing \begin{align*} |a_{n+1} - a_{n}| &= \left\lvert(-1)^{n+1} + \frac{1}{n+1} - (-1)^{n} - \frac{1}{n}\right\rvert \\ &= \left\lvert2(-1)^{n+1} - \frac{1}{n(n+1)}\right\rvert \\ &\geq 2 - \frac{1}{n(n+1)} \\ &\geq 3/2 \end{align*} for all $n \geq 1$.

If we know about Cauchy sequences and the fact that a convergent sequence is Cauchy, we can deduce that our sequence is not Cauchy, therefore not convergent.

Alternatively, fix an arbitrary real number $L$ as prospective limit and pick an arbitrary positive $\varepsilon < 3/4$. If $(a_{n}) \to L$, then for sufficiently large $n$ we have $|a_{n} - L| < \varepsilon < 3/4$. The triangle inequality implies $$ |a_{n+1} - a_{n}| \leq |(a_{n+1} - L) - (a_{n} - L)| \leq |a_{n+1} - L| + |a_{n} - L| < 3/4 + 3/4 = 3/2. $$ Contrapositively, since "$|a_{n+1} - a_{n}| < 3/2$" is false for all $n \geq 1$, the hypothesis $(a_{n}) \to L$ is false. Since $L$ was an arbitrary real number, $(a_{n})$ has no limit.

Answer 2

You've shown, perhaps, that neither $\ell = 1$ nor $\ell = -1$ (by symmetry) can be the limit. But could the sequence have a limit of $11.25$? You haven't shown that that is impossible. Nor have you addressed any of the tons of other possible limits. So you're not done.

Here's a suggestion: show that if the sequence $a_1, a_2, \ldots$ has a limit $L$, then any (infinite) subsequence also has the limit $L$.

Now see whether you can apply that idea to this problem.