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Evaluate $$\lim\limits_{n\to\infty} \frac{3+\sqrt{3}+\sqrt[3]{3}+\dots+\sqrt[n]{3}-n}{\ln n}.$$

We haven't been taught how to do asymptotes, I think this problem should be solved using Cesaro-Stolz; $$\lim\limits_{n\to\infty} \frac{a_n}{b_n}=\lim\limits_{n\to\infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n}$$ Note that this is my second question on this site.It's the first year i learned about limits so don't expect that this problem solves in a hard way.

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Yes, you are correct, we can use Stolz-Cesaro theorem: $$\lim_{n\to\infty} \frac{\sum_{k=1}^n(\sqrt[k]{3}-1)}{\ln(n)}\stackrel{SC}{=}\lim_{n\to\infty} \frac{\sqrt[n+1]{3}-1}{\ln (n+1)-\ln(n)}= \lim_{n\to\infty} \frac{e^{\left(\frac{\ln(3)}{n+1}\right)}-1}{\ln\left(1+\frac{1}{n}\right)}.$$ Now note that (see Arnaud Mortier's comment below) $$\lim_{t\to 0}\frac{e^{t}-1}{t}=\lim_{t\to 0}\frac{e^{t}-e^0}{t-0}=(e^t)'_{t=0}=e^0=1$$ and $$ \lim_{t\to 0}\frac{\ln(1+t)}{t}=1=\lim_{t\to 0}\frac{\ln(1+t)-\ln(1+0)}{t-0}=(\ln(1+t))'_{t=0}=\frac{1}{1+0}=1.$$ Can you take it from here?

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  • $\begingroup$ what do you mean by $\exp(\ln(3)/n)-1$ , btw i understand the denumerator $\endgroup$ – Robert Jan 15 '18 at 15:34
  • $\begingroup$ i never used that relation before, i can't do it, and btw $\lim_n \ln(1+\frac{1}{n})=0 ?$ $\endgroup$ – Robert Jan 15 '18 at 15:39
  • $\begingroup$ $$\lim_{t\to 0}\frac{e^{t}-1}{t}=1\quad\mbox{and}\quad \lim_{t\to 0}\frac{\ln(1+t)}{t}=1.$$ $\endgroup$ – Robert Jan 15 '18 at 15:44
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    $\begingroup$ You can find the final result by using $\lim_{t\to 0}\frac{e^{t}-1}{t}=1$ and $\lim_{t\to 0}\frac{\ln(1+t)}{t}=1$. $\endgroup$ – Robert Z Jan 15 '18 at 15:53
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    $\begingroup$ The two relations you mention can be seen as the definition of the derivative applied to the maps $e^t$ and ln$(1+t)$ at $0$. $\endgroup$ – Arnaud Mortier Jan 15 '18 at 15:53
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I thought it might be of interest to present an approach that does not rely on the Stolz-Cesaro Theorem, but rather uses only (i) a pair of inequalities of the exponential function, (ii) a result from the integral test, and (iii) the squeeze theorem. To that end, we present two primers with references to their proofs.


PRIMER $(1)$:

In THIS ANSWER, I showed using on the limit definition of the integral test and Bernoulli's Inequality that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le\frac{1}{1-x}} \tag1$$

for $x<1$.


PRIMER $(2)$:

As a result from the integral test (SEE THIS), we have for a function $f(x)$ that is positive valued and monotonically decreasing for $x\ge N$

$$\bbox[5px,border:2px solid #C0A000]{\int_N^{n+1}f(x)\,dx\le \sum_{k=N}^n f(k)\le f(N)+\int_N^n f(x)\,dx}\tag2$$


PROBLEM SOLUTION:

Equipped with Primers $(1)$ and $(2)$, we are ready to proceed.

We begin with the function $f(x)=\sqrt[x]{3}-1=e^{\frac1x\log(3)}-1$ and write the sum of interest as $\sum_{k=1}^n f(k)=f(1)+\sum_{k=2}^n f(k)=2+\sum_{k=2}^n f(k)$.

Inasmuch as $\lim_{n\to \infty}\frac{2}{\log(n)}=0$, we analyze the limit $\lim_{n\to \infty}\frac{\sum_{k=2}^nf(k)}{\log(n)}$.

Applying the left-hand sides of $(1)$ and $(2)$, we see that

$$\begin{align} \sum_{k=2}^nf(k)&\ge \int_2^{n+1}f(x)\,dx\\\\ &=\int_2^{n+1}(e^{\frac1x \log(3)}-1)\,dx\\\\ &\ge \int_2^{n+1}\frac{\log(3)}{x}\,dx\\\\ &=\log(3)(\log(n+1)-\log(2))\tag3 \end{align}$$

Applying the right-hand sides of $(1)$ and $(2)$, we see that

$$\begin{align} \sum_{k=2}^nf(k)&\le (\sqrt3-1)+\int_2^{n}f(x)\,dx\\\\ &=(\sqrt 3-1)+\int_2^{n}(e^{\frac1x \log(3)}-1)\,dx\\\\ &\le (\sqrt 3-1)+\int_2^{n}\frac{\log(3)}{x-\log(3)}\,dx\\\\ &=(\sqrt 3-1)+\log(3)(\log(n-\log(3))-\log(2-\log(3)))\tag4 \end{align}$$

Dividing the expressions in $(3)$ and $(4)$ by $\log(n)$, and applying the squeeze theorem, we obtain the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\frac{\sum_{k=1}^n(\sqrt[k]{3}-1)}{\log(n)}=\log(3)}$$

And we are done!

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    $\begingroup$ Very nice. The quality of contents and presentation of your answer is impressive. (+1) $\endgroup$ – Markus Scheuer Jan 15 '18 at 21:55
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    $\begingroup$ @MarkusScheuer Thank you for the nice words Markus! Very appreciative. $\endgroup$ – Mark Viola Jan 15 '18 at 22:11
  • $\begingroup$ I second the comment from @MarkusScheuer +1. IMHO this seems to be novel use of the idea behind integral test. $\endgroup$ – Paramanand Singh Jan 16 '18 at 16:22
  • $\begingroup$ @paramanandsingh Thank you my friend. And Happy New Year to you. $\endgroup$ – Mark Viola Jan 16 '18 at 16:32
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As noted already you are correct and by Stolz-Cesaro theorem: $$\lim_{n\to\infty} \frac{\sum_{k=1}^n(\sqrt[k]{3}-1)}{\ln(n)}\stackrel{SC}{=}\lim_{n\to\infty} \frac{\sqrt[n+1]{3}-1}{\ln\left(1+\frac{1}{n}\right)}=\ln 3$$

indeed

$$\frac{\sqrt[n+1]{3}-1}{\ln\left(1+\frac{1}{n}\right)}= \frac{n\left(\sqrt[n+1]{3}-1\right)}{\ln\left(1+\frac{1}{n}\right)^n} \to \ln 3$$

since by definition of $e$

$$\ln\left(1+\frac{1}{n}\right)^n\to \ln e=1$$

and by Taylor's espansion

$$\sqrt[n+1]{3}=e^{\frac{\ln 3}{n+1}}=1+\frac{\ln 3}{n+1}+o\left(\frac1n\right)$$

$$n\left(\sqrt[n+1]{3}-1\right)=\frac{n}{n+1}\ln 3+o(1)\to\ln 3$$

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