Limit of S_n, where $S_n = \int_{0}^{1}\frac{nx^{n-1}}{1+x} dx$ for $n\ge 1$

Question

$$S_n = \int_{0}^{1}\frac{n x^{n-1}}{1+x} dx \quad n \ge 1$$

My attempt:

$$|S_n| \le \int_{0}^{1}\left|\frac{n x^{n-1}}{1+x} dx\right| = \int_{0}^{1}\frac{n x^{n-1}}{1+x} dx \le \int_{0}^{1}n x^{n-1} dx$$ $0 \le x \le 1$, and $n \ge 1$.

So, $|S_n|=[x^n]_{0}^{1}$ and thus $|Sn| = 1$.

So, $\lim_{n \to \infty} S_n = 1$.

But the answer mentioned is $0$. I don't know what I am doing wrong. Please help.

Answer 1

Integrate by parts:

$$\int_0^1 {n x^{n-1} \over 1 + x}\, dx = {1 \over 2} + \int_0^1 {x^n \over (1 + x)^2}\,dx$$ Since $(1 + x)^2 \geq 1$, the integral on the right is bounded by ${\displaystyle\int_0^1 x^n\,dx = {1 \over n+1}}$. Hence the integral on the right goes to zero as $n$ goes to infinity and the limit of the left-hand integral will just be ${\displaystyle{1 \over 2}}$.

Answer 2

Your mistake is concluding $|S_n|=1$, when you have only shown that $|S_n| \leq 1$.

Answer 3

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ One possible method is to evaluate the integral. Then, we can take the limit of a known function. Indeed, as shown below, the integral is reduced to $$ \int_{0}^{1}{x^{n - 1} \over 1 + x}\,\dd x = \half\bracks{% \Psi\pars{n + 1\over 2} - \Psi\pars{n \over 2}} $$ where $\ds{\Psi}$ is the Digamma Function.

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\begin{align}&\color{#66f}{\large% \lim_{n\ \to\ \infty}\ \int_{0}^{1}{n x^{n - 1} \over 1 + x}\,\dd x} =\lim_{n\ \to\ \infty}\ \bracks{n\int_{0}^{1}{x^{n - 1} - x^{n} \over 1 - x^{2}} \,\dd x} \\[5mm]&=\lim_{n\ \to\ \infty}\ \bracks{n\int_{0}^{1}% {x^{\pars{n - 1}/2} - x^{n/2} \over 1 - x}\,\half\,x^{-1/2}\,\dd x} =\half\,\lim_{n\ \to\ \infty}\ \bracks{n\int_{0}^{1}% {x^{\pars{n/2 - 1}} - x^{n/2 - 1/2} \over 1 - x}\,\dd x} \\[5mm]&=\half\,\lim_{n\ \to\ \infty}\ \braces{n\bracks{% \int_{0}^{1}{1 - x^{n/2 - 1/2} \over 1 - x}\,\dd x -\int_{0}^{1}{1 - x^{n/2 - 1} \over 1 - x}\,\dd x}} \\[5mm]&=\half\,\lim_{n\ \to\ \infty}\ \braces{n\bracks{% \Psi\pars{n + 1\over 2} - \Psi\pars{n \over 2}}}\tag{1} \\[5mm]&=\half\,\lim_{n\ \to\ \infty}\ \braces{n\bracks{% \ln\pars{n + 1 \over 2} - {1 \over n + 1} - \ln\pars{n \over 2} + {1 \over n}}} \tag{2} \\[5mm]&=\half\,\lim_{n\ \to\ \infty}\ \braces{n\bracks{% \ln\pars{1 + {1 \over n}} + {1 \over n\pars{n + 1}}}} =\color{#66f}{\large\half} \end{align}

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$\ds{\Psi}$ is the Digamma Function.

In expression $\pars{1}$, we used the identity ${\bf 6.3.22}$.

In line $\pars{2}$, we used the Digamma Function Asymptotic Formula ${\bf 6.3.18}$.