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Let $a_n=\sqrt{n}$ and $s_n=a_1+a_2+\ldots+a_n$. Find the limit : $$\lim_{n\to +\infty}\left[\frac{\frac{a_n}{s_n}}{-\ln(1-\frac{a_n}{s_n})}\right]$$

I am hopelessly confused with this problem. Is L'Hopital or Cesaro-Stolz of any use here? Any hints. Thanks beforehand.

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3 Answers 3

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We have that by Stolz-Cesàro:

$$ \lim_{n\to +\infty} \frac{a_n}{s_n}=\lim_{n\to +\infty} \frac{a_{n+1}-a_{n}}{s_{n+1}-s_{n}}=\lim_{n\to +\infty} \frac{a_{n+1}-a_{n}}{a_{n+1}}=\lim_{n\to +\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}}=0 $$

Therefore:

$$ \lim_{n\to +\infty}\frac{\frac{a_n}{s_n}}{-\ln\left(1-\frac{a_n}{s_n}\right)}=\lim_{x\to 0}\frac{x}{-\ln\left(1-x\right)}=1 $$

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Observe that $x_n=a_n/s_n$ converges to zero. Then, the limit becomes trivially solvable by Taylor expansion of the logarithm.

Estimate:

$$s_n=\sum_{k=0}^n \sqrt{n}\approx \int_{0}^n \sqrt{n}dn=\frac{2}{3}n^{3/2}$$

(we are only interested in the power scaling - we need to see that $3/2>1/2$, so we don't need to talk about the error, but you can estimate the error from Euler-Maclaurin formula, which then proves that all the extra terms are negligible in the limit.

So you have

$$x_n=\frac{a_n}{s_n}\asymp\frac{1}{n}\underset{n\to \infty}{\to} 0 $$

By the way, you don't have to use integrals to show $a_n/s_n\to 0$, I just find it elegant. You can use induction or upper bounds or anything similar.

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  • $\begingroup$ thanks, so the final limit is $1$, by the L'Hopital rule. Can we use L'Hopital here $\endgroup$
    – vidyarthi
    Nov 29, 2018 at 12:02
  • $\begingroup$ Yes, after you reduce this to $\lim_{x\to0} \frac{x}{-\ln (1-x)}$, you can use whatever you want. Once you get some practice with these things, you remember $\ln(1+x)\approx x$ as one of the 5 most common approximations that you recognize in an instant, avoiding the need to calculate anything. $\endgroup$
    – orion
    Nov 29, 2018 at 12:08
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Observe that $$\lim_{t\to 0} \frac{t}{-\ln (1-t)} =\lim_{t\to 0} \frac{1}{(1-t)^{-1}} =\lim_{t\to 0} (1-t)=1$$ So by Heine definition of limit we have that for any sequence $x_n $ tending to $0$ the following equality holds $$\lim_{n\to \infty} \frac{x_n}{-\ln (1-x_n)}=1$$

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