Evaluate $\lim _{n\to \infty }n\int _1^2 \frac{dx}{x^2(1+x^n)}$

Question

Evaluate $$\lim _{n\to \infty }n\int _1^2 \frac{dx}{x^2(1+x^n)}$$

without Taylor expansion.

I tried rewriting as

$$\lim_{n\to \infty} n\int _1^2 \frac{1 + x^n - x^n}{x^2(1+x^n)}dx = \lim_{n\to \infty} n\int _1^2 \frac{dx}{x^2} - \lim_{n\to \infty} n\int _1^2 \frac{x^n}{x^2(1+x^n)}dx$$

The first integral is computable, but I don't know how to continue solving the second one.

The answer is $\ln 2$

the first integral is computable but the first limit diverges -- which is why it is not a good idea to break it up into 2 diverging parts — gt6989b, Apr 20 '17 at 19:02
@gt6989b Yes... I also thought about the substitution $t = \frac{1}{x}$ but it seems that doesn't help either, — Liviu, Apr 20 '17 at 19:12
@Dr.SonnhardGraubner I have the answer for the problem, I only have to solve it — Liviu, Apr 20 '17 at 19:18
@Dr.SonnhardGraubner: because $\int_{1}^{+\infty}\frac{dz}{z(z+1)}=\log(2)$. — Jack D'Aurizio, Apr 20 '17 at 19:26
That $\lim (A+B) = (\lim A) + (\lim B)$ is valid if $\lim A$ and $\lim B$ are finite numbers. But since $A$ is $n$ times a positive number not depending on $n,$ you have $\lim A=\infty$, so $\lim (A+B) = (\lim A) + (\lim B)$ is not valid in this case. — Michael Hardy, Apr 20 '17 at 19:34
However, it is perfectly valid to write $$ \lim_{n\to\infty} n \int_1^2 \frac{dx}{x^2(1+x^n)} = \lim_{n\to\infty} n\cdot\left( \int_1^2 \frac{dx}{x^2} - \int _1^2 \frac{x^n}{x^2(1+x^n)}dx \right), $$ with just one $\text{“}{\lim}\text{''}$ on the right side. — Michael Hardy, Apr 20 '17 at 19:43

score 3 · Accepted Answer · answered Apr 20 '17 at 19:23

3

By setting $x=z^{1/n}$ we are left with $$ \lim_{n\to +\infty}\int_{1}^{2^n}\frac{dz}{z(1+z)z^{1/n}}=\lim_{n\to +\infty}\left[O\left(\frac{1}{2^n}\right)+\int_{1}^{+\infty}\left(\frac{1}{z}-\frac{1}{z+1}\right)\frac{dz}{z^{1/n}}\right] $$ then by applying the dominated convergence theorem we get that the answer is $\color{red}{\log 2}$.

answered Apr 20 '17 at 19:23

Jack D'Aurizio

353,855

I haven't learned about the dominated convergence at my math classes... Could you, please, explain what to do next or what that means and why can it be applyed in this case ? – Liviu Apr 20 '17 at 19:30
1

@Liviu: the application is straightforward: https://en.wikipedia.org/wiki/Dominated_convergence_theorem – Jack D'Aurizio Apr 20 '17 at 19:33
1

The sequence of positive and integrable functions $f_n(z)=\left(\frac{1}{z}-\frac{1}{z+1}\right)\frac{1}{z^{1/n}}$ on the interval $(1,+\infty)$ is pointwise convergent to $\frac{1}{z}-\frac{1}{z+1}$ and in a monotonic way. – Jack D'Aurizio Apr 20 '17 at 19:35

score 1 · Answer 2 · answered Apr 21 '17 at 02:07

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\lim_{n \to \infty}\bracks{n\int_{1}^{2}{\dd x \over x^{2}\pars{1 + x^{n}}}} = \lim_{n \to \infty}\bracks{n\int_{1/2}^{1}{x^{n} \over 1 + x^{n}}\,\dd x} = \lim_{n \to \infty}\bracks{\int_{1/2^{n}}^{1}{x^{1/n} \over 1 + x}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\bracks{\int_{0}^{1}{x^{1/n} \over 1 + x}\,\dd x - \int_{0}^{1/2^{n}}{x^{1/n} \over 1 + x}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\bracks{\int_{0}^{1}{\dd x \over 1 + x} - \int_{0}^{1}{1 - x^{1/n} \over 1 + x}\,\dd x - \int_{0}^{1/2^{n}}{x^{1/n} \over 1 + x}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\bracks{\ln\pars{2} - H_{1/n} - \int_{0}^{1/2^{n}}{x^{1/n} \over 1 + x}\,\dd x}\qquad \pars{~H_{z}:\ Harmonic\ Number.\ \mbox{Note that}\ H_{0} = 0~} \end{align}

Moreover, $\ds{0 < \int_{0}^{1/2^{n}}{x^{1/n} \over 1 + x}\,\dd x < \int_{0}^{1/2^{n}}{x^{1/n} \over 1 + 0}\,\dd x = {n \over n + 1}\,{1 \over 2^{n + 1}}\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\, {\large 0}}$

$$ \bbx{\ds{\lim_{n \to \infty} \bracks{n\int_{1}^{2}{\dd x \over x^{2}\pars{1 + x^{n}}}} = \ln\pars{2}}} $$

Evaluate $\lim _{n\to \infty }n\int _1^2 \frac{dx}{x^2(1+x^n)}$

2 Answers2