I should calculate the limit of a sequence. A friend told me, that the solution is $1$. But I got $(-1)^n$.

The exercise is: $\lim\limits_{n \to \infty} \frac{1}{n^2} + (-1)^n \cdot \frac{n^2}{n^2+1}$

I did following: $$\begin{align*} &=\frac{n^2 ((-1)^n n^2 + 1 + \frac{1}{n^2})}{n^2(n^2+1)}\\ &=\frac{(-1)^n n^2 + 1 + \frac{1}{n^2}}{(n^2+1)}\\ &=\frac{n^2(\frac{(-1)^n n^2}{n^2} + \frac{1}{n^2} + \frac{1}{n^4})}{n^2(1 + \frac{1}{n^2})}\\ &=\frac{(-1)^n + 0 +0}{1}\\ &=\lim\limits_{n \to \infty} (-1)^n \end{align*}$$

What did I wrong?

Edit Well, some answers confused me. Here the complete exercise. I should check if the sequence is convergent for ${n \to \infty}$ and determine the limit if it exist. Also for a sequence which is $\infty$ or $-\infty$.

My friend got $1$ as limit. I got $(-1)^n$. I would say, that this sequence has no limit, just limit points $1$ and $-1$.

  • 1
    $\begingroup$ Nothing. (The limit doesn't exist.) $\endgroup$
    – only
    Aug 27, 2012 at 6:31
  • $\begingroup$ I did some tweaks to your TeX. In particular, please don't use $*$ for multiplication! This has a different meaning in mathematics (usually, convolution). Use \cdot instead. $\endgroup$ Aug 27, 2012 at 6:38
  • 4
    $\begingroup$ Your friend is wrong. $\endgroup$
    – JSCB
    Aug 27, 2012 at 7:50

3 Answers 3


The problem here is that the sequence does not converge at all!

To prove this, consider the subsequence where $n$ is an even number, and show that the limit is $1$. Then take the subsequence where $n$ is an odd number, and show that the limit is $-1$. Now, if a sequence converges, so do all its subsequences, and the limit is the same! Therefore it can't converge.

  • $\begingroup$ Ok, if I understand you correctly, the limit does not exist. Because the sequence is +1 -1 +1 -1 ... In german it is called Alternierende Reihe, is that correct? So my friend was wrong and my solution is correct? $\endgroup$
    – user23053
    Aug 27, 2012 at 6:47
  • $\begingroup$ @Taz: I assume you forgot commas in your comment, that is, you meant $+1,-1,+1,-1,\dots$. If so, you are mostly right: While the sequence is not that sequence, it "converges to" that sequence. It is indeed an alternating sequence (except for the first few terms), however the German "alternierende Reihe" means "alternating series"; the correct term in German would be "alternierende Folge" (note the summation sign present in the article you linked, which is absent in your problem). But apart from terminology, you are right and your friend is wrong. $\endgroup$
    – celtschk
    Aug 27, 2012 at 7:38
  • $\begingroup$ However note that being alternating is not sufficient for a limit to not exist (this is both true for series and for sequences). For example, consider $a_n=(-1)^n/n$. That sequence is clearly alternating, but it converges to $0$. Indeed, even the corresponding series converges (to $-\ln 2$), while the corresponding non-alternating series doesn't (but the non-alternating sequence of course does). $\endgroup$
    – celtschk
    Aug 27, 2012 at 7:44
  • $\begingroup$ @hofmeister One more point you seem to be missing: You can't say that the sequence converges to $(-1)^n$ simply because its dependent on $n$! But with respect to your intuition you are completely right. The subsequences converge to either $+1$ or $-1$! But it cant converge to both since the limit of a sequence is unique. The way to prove this was (as in my answer) to find two convergent subsequences with different (but unique) limits. To see what i mean you could look at the sequence $a_n = (-1)^n$ first! $\endgroup$
    – vanguard2k
    Aug 27, 2012 at 7:46

The limit of a sequence indexed by $n$ cannot contain $n$, as a limit is a number, not a sequence. So your answer $(-1)^n$ as outcome of the limit cannot be correct. However by writing $$ \frac1{n^2} + (-1)^n \frac{n^2}{n^2+1}=\frac1{n^2} + (-1)^n -(-1)^n\frac1{n^2+1} $$ you can see that you sequence is the sum of three sequences $\frac1{n^2}$, $(-1)^n$, and $-(-1)^n\frac1{n^2+1}$, of which the first and the last are convergent (to $0$), and the middle one is divergent, which implies that the sequence diverges (it has no limit).

  • $\begingroup$ Fun fact: It can contain $n$ if it's allowed as a letter, like $\sin$ :P Better wording: Cannot depend on $n$. $\endgroup$
    – AlexR
    Oct 8, 2014 at 18:06

Note that, for the limit of a sequence to exist, it has to be unique. That means the sequence has to converge to a single value. In your case the sequence has subsequences which converge to two different values, namely $ L = \left\{-1,1\right\}\,.$ So the limit does not exist. Here comes the notion of $\limsup$ and $\liminf$ which we use to define the existence of the limit. We say the sequence $a_{n}$ converges to $a$ $n\rightarrow \infty$, if

$$ \liminf_{n\rightarrow \infty} a_n =\limsup_{n \rightarrow \infty} a_n = \lim_{n \rightarrow\infty} a_n \,.$$

In your case $\liminf_{n\rightarrow \infty} a_n =-1$ and $ \limsup_{n \rightarrow \infty} a_n = 1 $, so they are not equal and hence the limit of the sequence does not exist.

You can compare this to the case of taking limits of functions when we require for the limit to exist at a point $x_0$ that $$ \lim_{x\rightarrow x_0^{-}} f(x) = \lim_{x\rightarrow x_0^{+}} f(x) = \lim_{x\rightarrow x_0} f(x) \,.$$

  • $\begingroup$ The comparison to limits of functions from the left and from the right is misleading, instead one could invoke $\limsup\limits_{x\to x_0}f(x)$ and $\liminf\limits_{x\to x_0}f(x)$. $\endgroup$
    – Did
    Aug 27, 2012 at 8:19
  • $\begingroup$ @SerialDownvoter:What's the story. $\endgroup$ Sep 9, 2012 at 17:06
  • $\begingroup$ I didn't downvote, but I agree with Did. $\endgroup$
    – Pedro
    May 29, 2013 at 0:42
  • $\begingroup$ @PeterTamaroff: I expect the OP know left and right limits of functions because usually they get expose to them early. So, that was the reason I wanted them to compare because the whole idea is to understand what the existence of limit mean. Otherwise, the answer is totally correct and there is no reason to be downvoted. $\endgroup$ Jun 28, 2013 at 22:09
  • $\begingroup$ @downvoter: the answer is very clear and to the point, so there is no need for the down vote and making argument for nothing. $\endgroup$ Dec 1, 2013 at 0:11

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