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I want to show that the sequence $$a_n=\frac{1}{n}+(-1)^n$$ does not converge to a limit.

I know that if a sequence $\left( a_n\right)_{n \in \mathbb{N}}$ converges to a limit L, then $\forall\epsilon >0\;\;\exists N\in\Bbb N \;\text{such that}\;\forall n\geq N, \;\;|a_n-L|\leq\epsilon$, but I am not sure how to use this to prove a sequence that does not converge to a limit.

Please give me some suggestions on where can I get started and how to write out the proof, thanks to anyone who can help.

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    $\begingroup$ check the limit for $n$ odd and $n$ even. $\endgroup$
    – Arian
    Oct 19, 2014 at 16:06
  • $\begingroup$ @Lucy: Your sequence is a sum of two parts. How each one of them behaves as $n$ tends to infinity? How will their sum behave? $\endgroup$
    – posilon
    Oct 19, 2014 at 16:13
  • $\begingroup$ You can easily show that $a_{n+1} - a_n$ goes to 2. If the sequence converged it would have been 0. $\endgroup$
    – lakshayg
    Oct 19, 2014 at 16:49

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Yo can prove that $a_n$ isn't Cauchy sequence. According the definition $a_n$ is a Cauchy sequence if:

$$\forall \varepsilon > 0 \, \exists N \in \mathbb N \, \forall n,m \geq N \; |a_n-a_m|<\varepsilon$$

But for $\varepsilon=\frac{1}{2}$ for all $N \in \mathbb{N}$ the difference between $a_N$ and $a_{N+1}$ is:

$$\left| a_N-a_{N+1} \right|= \left| \frac{1}{N}+(-1)^{N}-\frac{1}{N+1}-(-1)^{N+1} \right|=\left| 2 \cdot (-1)^{N}+\frac{1}{N}-\frac{1}{N+1} \right| \geq \\ \geq \left| 2 \cdot (-1)^{N} \right| - \left| \frac{1}{N}-\frac{1}{N+1} \right| \geq 2 - \frac{1}{N} - \frac{1}{N+1} \geq \frac{1}{2}$$

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Suppose $\;\lim a_n=L\;$ exists, then

$$(-1)^n=a_n-\frac1n\implies \lim_{n\to\infty}(-1)^n\stackrel{\text{Arith. of limits}}=\lim_{n\to\infty}a_n-\lim_{n\to\infty}\frac1n=L-0=L$$

But it is clear that $\;\{-1,1,-1,1,\;\ldots\}\;$ doesn't converge...

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  • $\begingroup$ Would you please explain the workings a bit more? $\endgroup$
    – Lucy
    Oct 19, 2014 at 16:12
  • $\begingroup$ What isn't clear, @Lucy? The first equality in the second line follows from the definition of $\;a_n\;$, the following from arithmetic of limits... $\endgroup$
    – Timbuc
    Oct 19, 2014 at 16:13
  • $\begingroup$ Sorry, I was just confusing myself earlier. I understand now, thank you for your help. $\endgroup$
    – Lucy
    Oct 19, 2014 at 16:19
  • $\begingroup$ @Lucy , my pleasure . $\endgroup$
    – Timbuc
    Oct 19, 2014 at 16:23
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If a sequence $\{ a_{n} \}_{n = 1}^{\infty}$ converges to a limit $L$, then every subsequence must converge (can you prove this part on your own?), and the subsequences must all converge to the same limit (can you prove this part on your own?).

But it is easy to prove that the subsequence $\{ a_{2n} \}_{n = 1}^{\infty}$ of even indices converges to $1$ (can you do this on your own?) while the subsequence $\{a_{2n - 1} \}_{n = 1}^{\infty}$ of odd indices converges to $-1$ (can you do this part on your own?).

Because of this, the sequence does not converge since it has two subsequences that converge to different limits.

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The subsequences $(a_{2n})$ and $(a_{2n+1})$ have different limits so the sequence $(a_n)$ is divergent.

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Notice that, if we choose $\varepsilon=\frac{1}2$, we would be able to find some $N$ and $L$ such that every term of $a_n$ with $n>N$ would be within $\frac{1}2$ of $L$. This is impossible, since odd terms, for $n>2$ are always less than $-\frac{1}2$ and even terms are always greater than $\frac{1}2$. Therefore, if an interval of size $\frac{1}2$ contained every even term, it could not contain any odd term and vice versa - implying that the limit cannot exist.

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