2

So, the problem that we were solving was

$$\lim_{n\to\infty}\left(\frac{n}{n+1}\right)^n$$

To figure out whether the series converged or diverged, after simplification, I asked my professor whether finding the limit of the inside of the function to determine whether the inside function was divergent or convergent would be helpful. My logic was this: If it's divergent, infinity to the power of $n$ as $n$ approaches infinity is just infinity, and if it's between $-1$ and $1$ then it approaches a finite value, right? Meaning that, if we were to take the limit of the inside it would ultimately determine what the function did after taken to the power of n.

I was told that we simply couldn't do this but her explanation was a bit lackluster, it was basically "because I said so." Please tell me exactly how I'm wrong so I can better understand what I'm doing. Thank you!

Robert Howard
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Florasce
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  • I try to explain it here. The limit is different, but the reasoning is similar. – Jyrki Lahtonen Oct 19 '18 at 16:12
  • @JyrkiLahtonen That helped me too! Thanks to you and everyone else who responded to this. I appreciate it! – Florasce Oct 19 '18 at 17:18
  • Just like one can't write $\sqrt{a+b} =\sqrt{a} +\sqrt{b} $ there are things that one can or can't do with limits. It is better to learn limit laws along with their proofs to understand what is or isn't allowed with limits. It's rather unfortunate that unlike algebra, learning of calculus mostly begins with a total disregard for any rules. – Paramanand Singh Oct 19 '18 at 18:37
  • I think that your question to your teacher is very reasonable. If you recognize that the term between brackets approaches a value like $2$, then indeed taking this to the power $n$ results in divergence. On the other hand, if the term approaches $1/2$, taking it to the power $n$ yields convergence, namely to $0$. The problem is that if the term between brackets is somewhat larger or smaller then $+1$ or $-1$, the result is inconclusive. You then have to use more powerful methods to determine whether a limit exists. – M. Wind Oct 20 '18 at 00:45
  • @M.Wind That's pretty much the exact logic I was using, yeah! (I'm glad someone recognized it.. ;; ) I knew I couldn't find the true value that way but it at least gives me something in a "yes or no" / "converge or diverge" format. – Florasce Oct 21 '18 at 18:50
  • I am glad I could reassure you that there is nothing wrong with your reasoning ! In general, if a mathematician is presented a new equation and is asked for his/her opinion, he/she will first perform a quick visual scan. Just to see which terms are large and which are small. Only if this is inconclusive he/she will go the stage of doing the hard math. – M. Wind Oct 22 '18 at 00:55
  • @M.Wind Thank you, I feel much better. Some of the other comments didn't really answer the logical question (rather than the computational one), thanks for this ;; I feel waaaay better. – Florasce Oct 23 '18 at 03:08

4 Answers4

3

In general we can't pull a limit past a variable that occurs in the limiting operation, because that might change the result. Here's an example which is easier than the problem you're working on:

$$ \lim_{x \to 1}\ (x-1)^{x-1} $$

If we take the limit of the inside function first, we'll get $0$, but in fact, the correct answer is $1$.

Now it is true that you can slide a limit operation past a variable that doesn't occur in the limit:

$$ \lim_{x \to 0} (x^y) = (\lim_{x \to 0} x)^y = 0^y = 0. $$

And it's true that you can slide a limit operation past a continuous function:

$$ \lim_{x \to \infty} \log(1 + \frac{1}{x}) = \log\left(\lim_{x \to \infty} 1 + \frac{1}{x}\right) = \log(1) = 0. $$

So it's easy to get tempted to think you can slide limits past whatever you want, but in fact you'll get the wrong answer.

hunter
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  • for the answer to the actual problem you're working on, how easy or hard it is will depend on what you already covered in class -- have you seen the limit definition of $e$, for example? – hunter Oct 19 '18 at 16:08
  • So this means it's incorrect, even if it's only used to find out whether the entire thing diverges or converges rather than the exact value? And yes, I have. I'm not trying to find the exact value of the function, I'm just trying to get more into formal proofs. – Florasce Oct 19 '18 at 16:09
  • @NicoleCarter that's right. – hunter Oct 19 '18 at 16:10
  • I see what you mean, thanks! That's a way better explanation than what my professor gave. This thing in particular had messed me up on past homework problems, where I figured that if the limit of the inside diverged (and of course it's the summation from n=1 to infinity) then since it's some number to the power of n that the entire thing diverged if both did. Right conclusion, wrong explanation. – Florasce Oct 19 '18 at 16:13
1

I'm not entirely certain how you're arguing this, but I think you're observing that $$-1 \le \frac{n}{n+1} \le 1,$$ and so raising this sequence to the power of $n$, you have $$0 \le \left(\frac{n}{n+1}\right)^n \le 1^n = 1,$$ which you claim makes the limit of the sequence finite.

It almost works; the problem is that divergence to infinity isn't the only way a sequence can fail to converge. Another example is the sequence $\cos(\pi n / 2)$, which takes the values $1, 0, -1, 0, 1, 0, -1, 0, \ldots$. It fails to converge, but is still bounded.

If you want to show that a bounded sequence still has a limit, you can try using the monotone convergence theorem. In this case, with a bit of work, you can show the sequence is monotone decreasing, and therefore has a limit.

Theo Bendit
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  • Thanks for the detailed response! I'm trying to get into more formal proofs and I wanted to see how this proposition was wrong, the "almost right but not quite" idea makes me feel like I didn't get goof it entirely up - I see what I did. Thanks! – Florasce Oct 19 '18 at 16:19
1

The reason why you can't do this is that the argument you've applied can lead you to the wrong conclusion, even for determining whether the limit is finite or not. For example, in the following limit the "inside" thing tends to $1$: $$ \lim_{n\to\infty}\left(1+\frac1{\sqrt n}\right)^n $$ so you would conclude that the limit is finite. It turns out that the limit is infinity.

grand_chat
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0

Explaining to someone "exactly why" they are wrong is actually quite difficult, because often their mistake is a result of fuzzy/intuitive reasoning and so its hard to pin down "exactly" the mistake. You say "my logic is..." but this isn't rigorous logic in the mathematical sense; you just aren't using the definition of the limit correctly.

  1. Infinity is not a number. There is no such thing as "infinity to the $n$".
  2. The limit of a sequence $a_n$ can never be given by an expression that depends on $n$. This wouldn't make any sense.
  3. In general operations of this kind can be seen to be invalid by considering basic examples like e.g. $$ 1 = n/n = n \times \tfrac{1}{n}. $$ First take limit of $1/n$. Its zero. So you have $n \times 0$. And this is always zero, so the total limit is zero. OR how about first take limit of $n$ to get infinity. But now you have infinity times $1/n$ which is always infinity!?

4.You are essentially considering $$ b_{m,n} = \Bigl( \frac{m}{m+1}\Bigr)^n, $$ and saying "let's imagine taking the limit as $m \to \infty$ first and then taking the limit as $n \to \infty$. This sort of limit swapping is a classic issue in analysis that sometimes can be done and other times cannot.

SBK
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  • Yeah, I see where I messed up now. I also wasn't working with paper at the time of this questioning. Thank you for the detailed response and I'll try to be a little more direct with my next questions so it's easier to see where I goofed. – Florasce Oct 19 '18 at 16:16