0
$\begingroup$

Theorem is : Every convergent sequence is bounded. Then what about the sequence $e^{1/(n−1)}$?

This series is convergent as $n$ tends to infinity but not bounded. Where am I going wrong?

  • 0
    What is your approach? – 2017-02-05
  • 0
    A convergent sequence is always bounded. – 2017-02-05
  • 0
    What makes you believe that this is not bounded? – 2017-02-05
  • 0
    Also, there's a big different between 'series' and 'sequence'. – 2017-02-05
  • 0
    How do you conclude that it isn't bounded? I'm afraid I disagree. – 2017-02-05
  • 0
    the sequence starts from infinity at n=1. – 2017-02-05
  • 0
    Then your sequence is undefined in $n=1$, not starts from infinity. – 2017-02-05
  • 0
    Hey voters I say this as a complete hardass about questions, I really don't think this question deserves downvotes. It seems like there's an honest misunderstanding. The quality is kind of low (format/language issues) so I get if you downvote for that. I'm just saying that the content is a reasonable question. – 2017-02-05
  • 1
    Well, $1-1=-0$, therefore obviously $e^{1/(1-1)} = e^{-1/0} = e^{-\infty} = 0$. Find the error in my argumentation, and you have the error in yours. – 2017-02-05

5 Answers 5

2

There are two issues here:

  1. Your sequence does not "start from infinity at $n=1$". There is no such number. Your expression $e^{\frac{1}{n-1}}$ does not define a number if $n=1$. I cannot stress this enough.
  2. It doesn't sound like you have the definition of bounded right. A sequence is bounded if there is a fixed number $M$ such that every term of the sequence $a_{n}$ has $|a_{n}|M$

Notice now that the sequence you specify is bounded (for $n \ge 2$) by $e$.

1

$e^{1/(n-1)}$ is not defined for $n=1$. So the sequence must start at $n=2$. This has been covered in other answers so there's nothing to add.

However, you were mentioning that for $n=1$, the sequence is "infinity". Let investigate this. In various areas of math, sometimes it IS useful to have a notion of "infinity", we call this the extended real numbers. If you insist that $e^{1/(1-1)}=\infty$, then we still have that this sequence is bounded because then $\infty$ is a perfectly valid element. We have that $\infty\ge e^{1/(n-1)}\ge 0$.

However, this is definitely not standard. Most people would say that $e^{1/(n-1)}$ is just undefined.

  • 1
    Even in the extended real numbers it's undefined. It could either be $\infty$ or $0$, because there's no reason to pick $\infty$ instead of $-\infty$. – 2017-02-05
  • 0
    @MattSamuel Very good point. Thanks. I was just considering if he insisted that $a_1=\infty$ for whatever reason – 2017-02-05
0

Every convergent sequence is bounded. The sequence $(a_n)_{n \in \mathbf N_{\geq 2}}$ with $a_n = \mathrm e^\frac{1}{n-1}$ is bounded because $0 < \frac{1}{n-1} < 1$ for all $n\in \mathbf N_{\geq 2}$ implies $$1=\mathrm e^0 < a_n < \mathrm e$$ where we used that the exponential function is monotone.

  • 0
    Thanks for replying. Why aren't you taking n = 1 ? – 2017-02-05
  • 1
    The sequence is not well defined for $n=1$ since we would have $1/(n-1)=1/(1-1)=1/0$. – 2017-02-05
0

It's a common mistake to suppose that $1/0 = \infty$. $\infty$ is not a number; no numeric expression takes value $\infty$. To see one reason why, consider the function $f(x) = \frac{1}{x}$. Just after $0$, when $x$ is a small positive number, $f$ is very large, moving toward $\infty$. When $x$ is a small negative number, $f$ is very negative, moving toward $-\infty$. If we were to say $\frac{1}{0} = \infty$, we would have just as much reason to say $\frac{1}{0} = -\infty$, so $\infty = -\infty$, which is nonsense. $1/0$ is therefore simply undefined. Your sequence $e^{\frac{1}{n-1}}$ doesn't start from infinity and decrease into finite numbers, it starts with a "hole" where it's undefined. And a function $g(n)$ is "bounded" if there is some $M$ so that $g(n) \leq M$ whenever $g(n)$ is defined. So $e^{\frac{1}{n-1}}$ is indeed bounded, with upper bound $e$ (or, e.g., $e + 1$ if you want a strict upper bound).

0

There is one case where we are allowed to define $1/0$ as $\infty$, and that is in the Riemann sphere (extended complex plane) $\mathbb C\cup\{\infty\}$. We then also have $e^\infty=\infty$. No matter what metric you use though, every sequence in the Riemann sphere is bounded.