StWolfzfram's syndrome

Calculus Level pending

Compute $\large \displaystyle \lim_{n \to \infty} \frac {\pi \cdot {n}}{1 + \sqrt[2]{2} + \sqrt[3]{3} + \cdots + \sqrt[n]{n}}.$ Give your answer to 2 decimal places.

The answer is 3.14.

2 solutions

Otto Bretscher
Nov 24, 2015

Clever problem! I will have to assign this in a calculus exam some time.

$\sum_{k=1}^{n}\sqrt[k]{k}$ is asymptotically equal to $n$ since $\lim_{n\to\infty}\sqrt[k]{k}=1$ , so that the limit we seek is $\pi\approx \boxed{3.14}$

I have a doubt sir. when you say $\sum_{k=1}^{n}\sqrt[k]{k}$ is asymptotically equal to n I can only see it applying Stolz,are you applying Stolz Criterium? I'm not sure how do you do it

Guillermo Templado - 5 years, 6 months ago

No, just a simple $\epsilon$ -argument... all I'm saying is that the average value of $1,...\sqrt[n]{n}$ goes to 1 as we let $n$ go to infinity.

Otto Bretscher - 5 years, 6 months ago

thank you very much,althought anyway I only get understand you with Stolz. If noone posts a solution with Stolz, I'll post a solution with this one later.

Guillermo Templado - 5 years, 6 months ago

@Guillermo Templado – What I'm saying is this: If a decreasing sequence $a_n$ approaches a limit $a$ , then the average $b_n=\frac{\sum_{k=1}^{n}a_k}{n}$ will also approach $a$ ; that follows directly from the definitions.

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – thank you very much, I got it now, althought it's not necessary $a_{n}$ decreasing,is it? If $\lim_{n \to \infty} a_{n}$ = a , then $\lim_{n \to \infty} \frac {\sum_{k=1}^{n} a_{k}}{n}$ = a

Guillermo Templado - 5 years, 6 months ago

@Guillermo Templado – Nope. In Stolz lemma, only b_n is required to monotonically increase/decrease (so it diverge).

Pi Han Goh - 5 years, 6 months ago

@Pi Han Goh – Ah, right,thank you

Guillermo Templado - 5 years, 6 months ago

@Guillermo Templado – "decreasing" (or "increasing") is not logically necessary, but it makes the argument easier. I include the condition when I do problems like that in my calculus classes, as a special case of $\lim_{x\to\infty}\frac{1}{x}\int_{0}^{x}f(t)dt=\lim_{x\to\infty}f(x)$ if the last limit exists.

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – I don't usually question the comments or answers, because the only right answer must be found by oneself, sometimes I simply prefer to laugh... Thank you again, for your time this time, it wasn't neccesary. This isn't a clever question, you are a clever person ,it's different.

Guillermo Templado - 5 years, 6 months ago

ok,I think I'm understanding you now, considerating $\lim_{k \to \infty}\sqrt[k]{k}$ = 1

Guillermo Templado - 5 years, 6 months ago

What is $\epsilon$ -argument? Is it Epsilon delta?

Pi Han Goh - 5 years, 6 months ago

@Pi Han Goh – Well, there is no delta in a sequence, only an epsilon ;) For every $\epsilon>0$ there is an $n$ such that the average of $1,..,\sqrt[n]{n}<1+\epsilon$

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – Is this something obvious (the average is less than 1 + epsilon)? I don't think it's a common fact. You might want to prove it rigorously.

Pi Han Goh - 5 years, 6 months ago

@Pi Han Goh – It's pretty obvious for a decreasing sequence $a_n$ with limit $a$ , yes. Given any $\epsilon>0$ , there exists an $N$ such that $a_n<a+\frac{\epsilon}{2}$ whenever $n>N$ . By adding enough terms past the $N$ 'th, we can push the average below $a+\epsilon$ .

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – Hmmmm... Either I'm getting rusty in my calculus or you're setting too high of a standard for us.... I think it's most likely the latter case. HHAAHAHAHAHA

Pi Han Goh - 5 years, 6 months ago

@Pi Han Goh – I'm adding my answer one sentence at a time, since my "brilliant.org" often crashes. See the whole answer, please. Does it make more sense now?

Otto Bretscher - 5 years, 6 months ago

@Otto Bretscher – Ah that made sense. You should post that in your solution.... Or just use Stolz Cesaro Theorem hahahahaahhaha

Pi Han Goh - 5 years, 6 months ago

Guillermo Templado
Feb 2, 2016

* Stolz Criterium *

Let $\{a_{n} \}$ and $\{b_{n} \}$ two sequences such that:

$\lim_ {n \to \infty} a_{n} = 0$ or $\infty$ ,and $\{b_{n} \}$ is monotonically decreasing and $\lim_ {n \to \infty} b_{n}$ = 0 or monotone increasing and diverging to + $\infty$ . $\lim_ {n \to \infty} \frac {a_ {n + 1} -a_{n}}{b_ {n + 1} -b_{n}} = \lambda, \lambda \in \mathbb{R}$ Then the limit:

$\lim_ {n \to \infty} \frac {a_{n}} {b_{n}} = \lambda$ .

In this case

$\lim_{n \to \infty} \frac{\pi \cdot (n + 1) - \pi \cdot n}{\sqrt[n + 1]{n + 1}} = \lim_{n \to \infty} \frac{\pi}{\sqrt[n + 1]{n + 1}} = \pi \Rightarrow$ $\lim_{n \to \infty} \frac{\pi \cdot n}{\sqrt[1]{1} + \sqrt[2]{2} + \ldots + \sqrt[n]{n}} = \pi$

StWolfzfram's syndrome

The answer is 3.14.

2 solutions

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