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I'm trying to calculate: $$T = \lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2$$ Here is my attempt.

Put $x^2=\dfrac{1}{t}$ so when $x\to \infty, t \to 0$ and the limit become \begin{align*} T &= \lim\limits_{t \to 0} \sqrt[n]{\left(1+\dfrac{1}{t}\right)\left(2+\dfrac{1}{t}\right)...\left(n+\dfrac{1}{t}\right)}-\dfrac{1}{t}\\ &=\lim\limits_{t \to 0} \sqrt[n]{\left(\dfrac{t+1}{t}\right)\left(\dfrac{2t+1}{t}\right)...\left(\dfrac{nt+1}{t}\right)}-\dfrac{1}{t} \\ &=\lim\limits_{t \to 0} \dfrac{\sqrt[n]{(t+1)(2t+1)...(nt+1)}-1}{t} \end{align*} My idea is to use $\lim\limits_{x\to0}\dfrac{(ax+1)^{\beta}-1}{x} =a\beta .$ But after some steps (above), now I'm stuck. Thanks for any helps.

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    $\begingroup$ "Put $x^2=t$ so when $x\to \infty, t \to 0$" ... You meant $x^2 = 1/t $? $\endgroup$
    – leonbloy
    Jul 27, 2020 at 12:55
  • $\begingroup$ Thanks, it's my bad. Edited $\endgroup$
    – aDmaL
    Jul 27, 2020 at 12:58

6 Answers 6

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The idea is very good! The limit should be for $t\to0^+$, but since the limit for $t\to0$ exists, there's no real problem. However, you should use $t\to0^+$ for the sake of rigor.

The two-sided limit is the derivative at $0$ of the function $$ f(t)=\sqrt[n]{(t+1)(2t+1)\dotsm(nt+1)} $$ and in order to compute it, the logarithmic derivative is handy: $$ \log f(t)=\dfrac{1}{n}\bigl(\log(t+1)+\log(2t+1)+\dots+\log(nt+1)\bigr) $$ and therefore $$ n\frac{f'(t)}{f(t)}=\frac{1}{t+1}+\frac{2}{2t+1}+\dots+\frac{n}{nt+1} $$ which yields $$ n\frac{f'(0)}{f(0)}=1+2+\dots+n=\frac{n(n+1)}{2} $$ Since $f(0)=1$, we have $$ f'(0)=\frac{n+1}{2} $$

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  • $\begingroup$ Is there any other ways? Because sometimes (like this one), it's a bit hard to see it in form of a derivative of a function. $\endgroup$
    – aDmaL
    Jul 27, 2020 at 13:00
  • $\begingroup$ @aDmaL You can use the Taylor expansion as in Houa's answer, but it's essentially the same thing: you need to compute the coefficient of $t$, which is the derivative at zero of $(t+1)(2t+1)\dotsm(nt+1)$, and the logarithmic derivative is again handy. To the contrary, I find it easier to spot a derivative when the denominator is $t$, like in this case. $\endgroup$
    – egreg
    Jul 27, 2020 at 13:02
  • $\begingroup$ Thank you! Can you check out this Linear Algebra problem? I encountered this a few month ago, but still haven't fully understand the instruction. yet. Can you helo me? I think this problem can be generalize to $n$ functions of oder $n$ and $n$ parameters, though. math.stackexchange.com/questions/3454702/… $\endgroup$
    – aDmaL
    Jul 27, 2020 at 13:20
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Based on your way, in fact you can calculate $(1+t)(1+2t)...(1+nt)$ directly. $$(1+t)(1+2t)...(1+nt)=a_nt^n+...+\frac{(n+1)n}{2}t+1=f(t)+1$$ and $f(t)$ tends to $0$. So $$(f(t)+1)^{\frac{1}{n}}-1\sim\frac{1}{n}f(t)$$ and $$T=\lim_{t\rightarrow 0^+}\frac{1}{n}\frac{f(t)}{t}=\frac{n+1}{2}$$

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  • $\begingroup$ A bit ambiguity for me about the coefficient index. But it indeed easier to see than seen it in form of a derivative. Thank you! $\endgroup$
    – aDmaL
    Jul 27, 2020 at 13:20
  • $\begingroup$ @aDmaL Derivative, Taylor expansion, Infinitesimal equivalent, and Lagrange etc. are all useful for these "homework-like" problems. one should grasp them deeply. And in fact they have the same nature. For example $\frac{e^x-1}{x}$, you can see it as the derivative of $e^x$, or the first and second terms of Taylor expansion of $e^x$, and so on $\endgroup$
    – Houa
    Jul 27, 2020 at 13:24
  • $\begingroup$ Thanks for the advice! Sometimes these massive $f(t)$ make some "newbie" like me really struggle. So glad that always have you guys support me. Math is awesome, it's connect people. $\endgroup$
    – aDmaL
    Jul 27, 2020 at 13:31
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The limit can also be shown using HM-GM-AM.

Setting $u = x^2$ and considering $u\to +\infty$ we have

$$\frac n{\sum_{k=1}^n\frac 1{k+u}} - u \leq \sqrt[n]{\prod_{k=1}^n (k+u)} - u \leq \frac{\sum_{k=1}^n(k+u)}n - u = \frac{n+1}{2}$$

For the LHS we have

$$\frac n{\sum_{k=1}^n\frac 1{k+u}} - u = \frac{n - \sum_{k=1}^n\frac u{k+u}}{\frac 1u\sum_{k=1}^n\frac 1{\frac ku+1}} $$ $$= \frac{u \sum_{k=1}^n\frac k{k+u}}{\sum_{k=1}^n\frac 1{\frac ku+1}}= \frac{\sum_{k=1}^n\frac k{\frac ku+1}}{\sum_{k=1}^n\frac 1{\frac ku+1}}\stackrel{u\to+\infty}{\longrightarrow}\frac{\sum_{k=1}^n k}{n} = \frac{n+1}{2}$$

Now, squeezing gives the limit $ \frac{n+1}{2}$.

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By definition of pochammer symbol $$(x^2+1)^{(n)}=(1+x^2)(2+x^2)\cdots (n+x^2)=\frac{\Gamma(x^2+1+n)}{\Gamma(x^2+1)}\sim x^{2n}\left(1+\frac{n(n+1)}{2x^2}+O(x^{-4})\right)$$ thus $$\sqrt[n]{(x^2+1)^{(n)}}-x^2 =x^2\left(1+\frac{n(n+1)}{2x^2}\right)^{\frac{1}{n}}-x^2$$ using the fractional binomial theorem we have limit $$\lim_{x\to \infty}\left(\sqrt[n]{(x^2+1)^{(n)}}-x^2\right)= x^2\left(1+\frac{n(n+1)}{2n} x^{-2} +O(x^{-4}) -x^2\right)=\frac{n+1}{2}$$

Notation: $O(.)$ is Big O notation.

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Hint:

In the development of the product under the radical, the dominant terms are

$$x^{2n}+(1+2+\cdots n)x^{2n-2}=x^{2n}\left(1+\frac{n(n+1)}2x^{-2}\right).$$

Then, taking the $n^{th}$ root, ($\sqrt[n]{1+x}=1+\frac xn+\cdots$),

$$x^2\left(1+\frac{n(n+1)}{2n}x^{-2}+\cdots\right)-x^2\to\frac{n+1}2$$ as the other terms are of a lower order.

You can make it rigorous with the $o$ notation.

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$$\lim\limits_{x \to \infty} \sqrt[n]{(1+x^2)(2+x^2)...(n+x^2)}-x^2=\lim\limits_{x \to \infty}x^2\left[ e^{\frac{1}{n}\ln \left(1+ \frac{1}{x^2} \right)\cdots \left(1+ \frac{n}{x^2} \right)}-1 \right] =\\=\lim\limits_{x \to \infty}\frac{1}{n}x^2\left[ \ln \left(1+ \frac{1}{x^2} \right)+\cdots+\ln \left(1+ \frac{n}{x^2} \right)\right]=\frac{1}{n}\left[1+2+ \cdots+n\right] =\frac{n+1}{2} $$

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  • $\begingroup$ Ready to explain each step in case of doubt. Imho, easiest way. $\endgroup$
    – zkutch
    Jul 27, 2020 at 13:38
  • $\begingroup$ The exponential should be $\ln\left[\left(1+\dfrac{1}{x^2}\right)...\left(1+\dfrac{n}{x^2}\right)\right]$. At the next line is fine, maybe just some typos, though. A really wise way, thank you a lot. $\endgroup$
    – aDmaL
    Jul 27, 2020 at 13:43
  • $\begingroup$ Yes, of course. Thanks, fixed. And also thanks for estimating: I prefer to keep things as simple as possible. $\endgroup$
    – zkutch
    Jul 27, 2020 at 14:41

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