# Prove that the sequence $\{n + \frac{1}{n} \}^{\infty}_{n=1}$ has no limit

Attempt:

Suppose $\epsilon = 1$, and there exists a limit $L$,

$|n+1/n - L| = |\frac{n^2 +1}{n} - L|>1$ for all $n\geq N$ since $\frac{n^2 +1}{n}>0$

This seems to simple and I believe I am missing key step to show that in fact for infinite $n$ the definition for the limit of a sequence does not hold.

• Why not writing $N$ out explicitly? – Vim Jul 30 '16 at 1:08
• The fact that $\frac{n^2+1}{n}\gt 0$ is not enough. But let $N=\lceil |L|\rceil+88$. If $n\gt N$, then the absolute difference is $\gt \epsilon$. – André Nicolas Jul 30 '16 at 1:15
• @AndréNicolas Is the L in absolute value. – M. Ebsim Jul 30 '16 at 1:26
• Yes, I took the absolute value to simplify things. And then the ceiling function. As Vim indicated, you should be explicit in showing that there is no $N$ beyond which the inequality holds. I should not have used $N$ for my number, should have called it $K$. Since for any $n\gt K$ the inequality fails, there is no $N$ beyond which the inequality holds. Later one can be somewhat more casual, when it is clear that minor details can be filled in. But right now you are in the process of persuading yourself, and others, that you can deal with all details. – André Nicolas Jul 30 '16 at 1:34

If a sequence converges, it is bounded. Your sequence is not bounded, so it cannot converge.

• The place in which the question is placed presumes that I don't discuss bounded sequences. – M. Ebsim Jul 30 '16 at 1:10
• Well, then simply note the distance between any two $n + 1/n$ is at least $1$ so for any $1/2 > \epsilon > 0$ will have at most one term of the sequence in any neighborhood with diameter 1/2. So no number can be a limit. – fleablood Jul 30 '16 at 3:53

By way of contradiction: If the limit exists, call it $L$, then take $\epsilon = 0.1$ (according to the definition I can take ANY $\epsilon > 0)$, and find $N$ sufficiently large so that $| a_n - L | < \epsilon$ for all $n \ge N$. In particular, $| a_n - a_{n+1} | < 2 \epsilon = 0.2$ - but that is impossible, all differences are $\; \ge 0.5$ (say).

Note that if $m > n$ then $(m + 1/m) - (n+ 1/n) = (m-n) + (1/n - 1/m) > 1$.

So for any $x \in R$, $[x - 1/2, x + 1/2]$ contains at most one $n + 1/n$ point.

So there is at most one possible $n + 1/n$ such that $|(n + 1/n) - x| < 1/2$.

So it is not possible that there is an $M$ so that for all $n > M$ it follows $|(n + 1/n) - x| < 1/2$.

So $x$ can not be a limit point.

This is true of all $x \in R$ so {$n + 1/n$} has no limit points.