-1

$$\sum_{x=0}^{\infty} 0^x$$
Check whether the infinite series above converges or diverges.

Using the root test, the series will converge (as the limit is $0$).
Using the ratio test:
$\lim_\limits{x \to \infty} \Big|\frac{a_{n+1}}{a_{n}}\Big|$ = $\lim_\limits{x \to \infty} \Big|\frac{0^{x+1}}{0^x}\Big|$ =$\lim_\limits{x \to \infty} \Big|\frac{(0)0^{x}}{0^x}\Big|$ = $0$
And the series converges.

Evaluating the limit from the ratio test in a different manner:
$\lim_\limits{x \to \infty} \Big|\frac{a_{n+1}}{a_{n}}\Big|$ = $\lim_\limits{x \to \infty} \Big|\frac{0^{x+1}}{0^x}\Big|$ =$\lim_\limits{x \to \infty} \Big|\frac{0^{x}}{0^x}\Big|$ = $1$
And the ratio test is inconclusive.

It turns out that the limit in the ratio test is in the form $\frac{0}{0}$ and is indeterminate.
(If you take a look at the graph of $0^x$, it mostly consists of $0$s as $x$ approaches infinity.)

However, the series should diverge as $0^0$ is undefined. Is this a failure of the ratio test and root test?

helpme
  • 661
  • 1
    If $0^0$ is undefined then questions of divergence/convergence don't even make sense. It's a question of the question being well-defined in the first place. – Randall Nov 25 '19 at 13:31
  • Anyway, no, this is not a failure of any of these mathematical truths. These things say IF a limit exists and behaves a certain way, then you can conclude stuff. This doesn't mean you can conclude anything if the hypotheses are not met. – Randall Nov 25 '19 at 13:36
  • 4
    We have $0^0=1$, period. So clearly $\sum_{n=0}^\infty 0^n=1+0+0+\ldots=1$. However, for $n>0$ the quotient $\frac{a_{n+1}}{a_n}$ is undefined (division by $0$), not merely indeterminate – Hagen von Eitzen Nov 25 '19 at 13:36
  • @HagenvonEitzen why do we have $0^0=1$? – Randall Nov 25 '19 at 13:49
  • @Randall see https://en.wikipedia.org/wiki/Zero_to_the_power_of_zero and https://math.stackexchange.com/questions/11150/zero-to-the-zero-power-is-00-1 – JMoravitz Nov 25 '19 at 13:52
  • 1
    $0^x=0$ and story ends. – TheStudent Nov 25 '19 at 14:08
  • @TheStudent: $0^0$ is undefined and the story ends. (Following another convention, $0^0=1$ and the story ends.) –  Nov 25 '19 at 14:09
  • @JMoravitz the second link (which I liked reading) seems to agree that $0^0$ isn't necessarily $1$, period. (At least the accepted answer with 300+ upvotes.) – Randall Nov 25 '19 at 14:15
  • may be solution can be defined if approaching to $0^0$ form – TheStudent Nov 25 '19 at 14:27
  • @Randall and yet that same answer says that if we were to define it as anything other than undefined then the only logical choice is $1$. Really, it depends on context, but the contexts where $0^0=1$ is useful for are so prevalent that to many people they drop the qualifying statements and just flat-out say it equals $1$ unconditionally. Of course someone who is saying $0^0=1$ isn't going to make the mistake of saying that $\lim\limits_{(x,y)\to (0,0)}x^y$ also equals one, that clearly is undefined regardless of context or opinion. – JMoravitz Nov 25 '19 at 14:52

3 Answers3

1

Contrary to your belief,

  1. $0^0$ is meaningless, making any partial sum undefined,

  2. $\dfrac{0^{x+1}}{0^x}$ is meaningless (it is not $0$),

and all this discussion is meaningless.

What we can say is that

$$t_0+\sum_{n=1}^\infty 0^n=t_0+0+0+0+0+\cdots=t_0,$$ whatever you decide $t_0$ to be.

No convergence test is required.

  • This answer is nonresponsive to OP's question about potential failure of the ratio and root tests. If you wish to challenge the framing of the Question or to improve the content of the Question, comments to the Question are the correct venue. – Eric Towers Nov 25 '19 at 19:24
0

The value of $0^0$ is irrelevant. The limit as $x \rightarrow \infty$ only inspects large $x$. If it is possible to pick a bound $M$ such that for all $x > M$, $0^x$ is defined, (so, necessarily, $M > 0$), we may ignore all $x$ less than $M$.

Both of your applications of the ratio test include either the expression $\frac{0^n}{0^{n+1}}$ or the expression $\frac{0^{n+1}}{0^n}$. These expressions are undefined for all $n >1$. Consequently, the limit sees no sequence of values, so there is no limit. Relevant to the argument in this Answer's first paragraph, we may ignore $n= 0$ since, for convergence, a finite initial segment of the sequence may be ignored.

To summarize: A limit of expressions which persistently do not exist does not exist. This is not because a test succeeds or fails, but because limits are defined by values of various expressions. If those values do not exist, the limit cannot.

Eric Towers
  • 67,037
0

Note that for $x>0$ we have $0^x=0$ therefore

$$\frac{0^{x+1}}{0^x}=\frac 0 0$$

in an undefined expression and therefore we can take the limit and ratio test doesn't apply.

Starting from $x=1$, the sum is well defined and we simply have

$$\sum_{x=1}^{\infty} 0^x=\lim_{N\to \infty}\sum_{x=1}^{N} 0=\lim_{N\to \infty}(\overbrace{0+0+\ldots+0}^{\text{N terms}})= \lim_{N\to \infty} 0=0$$

For the issue on the starting value for $x\ge 0$, refer to

user
  • 154,566
  • Mh, what is the term $0^0$ ?? –  Nov 25 '19 at 14:06
  • @YvesDaoust Opssss, of course we need to start from 1! Thanks – user Nov 25 '19 at 14:14
  • Are you sure ?? –  Nov 25 '19 at 14:56
  • @YvesDaoust It depends if we want assing a value to $0^0$ or not, anyway what is relevant for a series is the behaviour for $x$ large, so I think we can also ignore the case $x=0$ and consider the sum starting from $x=1$. – user Nov 25 '19 at 15:00
  • This is wrong. The sum starts at $x=0$ and you may not drop this term, undefined is undefined. There is no consensus on $0^0=1$. Could you swear that $\Re ((-1)^\pi)+10^{1000000}>0$ ? –  Nov 25 '19 at 15:02
  • @YvesDaoust What I mean is that since the behaviour of a series is not dependent by as large we fix the starting value, to deteremine convergence or divergence, we can also exclude the starting values for which the expression is not well defined. Of course we can also assign a value to these undefined expression as you did, it is fine of course, but I don't agree with you taht it is the only way to approach the problem. – user Nov 25 '19 at 15:06
  • No, you may not exclude the starting values, this is a serious misconception. By the way, I did not assign a value to $0^0$, did I ?? –  Nov 25 '19 at 15:08
  • @YvesDaoust Ah ok now I get your point. Ok we can also say that $0^0$ is come $t_0$ unknown value, but then aren't you assigning $t_0$? Anyway, in this case I don't think the main point from tha asker is that one, even it is important to clarify. It seems to me that the main point is on the wrong application of ratio and root test and on the fact that we don't need any test to find the sum (for $x\ge 1$), as you properly have pointed out. – user Nov 25 '19 at 15:12
  • I wrote "whatever you decide $t_0$ to be" on purpose, I didn't assign any value. And if you decide that $t_0$ is undefined, my answer is still valid. –  Nov 25 '19 at 15:17
  • @YvesDaoust Assign a symbol to a undecided value is quite strange. It $0^0$ is undefined, isn't it meaningless denote it by $t_0$? Anyway I will change something in order to avoid any confusion with that. Thanks for your kind suggestion and interesting discussion. Bye – user Nov 25 '19 at 15:22
  • Ok, please replace all occurrences of $t_0$ by "He Who Cannot Be Named", in white letters. –  Nov 25 '19 at 15:24
  • @YvesDaoust Yes, indeed it is mainly a matter of language and notation. In ordert to help the asker on the issue for $0^0$ I've linked a specific OP on that. – user Nov 25 '19 at 15:26
  • The OP said "the series should diverge as $0^0$ is undefined", which leaves little doubt about his opinion on $0^0$. At the same time, his use of "diverge" is questionable. –  Nov 25 '19 at 15:42