9

I have got one method,

If we consider $ a_{n} = \int_{0}^{1} \frac{nx^{n-1}}{1+x} \ dx $

Then, $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) = \lim_{n \to \infty }a_{n} = \frac{1}{2} $

But can anyone attack this problem in a different & more standard way?

Arctic Char
  • 16,007
Mathematics Man
  • 353
  • I would like to make the following observation: since the alternating series inside the brackets naturally occurs as the remainder (so-called tail) of the alternating harmonic series, it is only natural to evaluate this remainder with the same tools of integral representation which enable one to find the sum of the said series. In other words, the method you have mentioned is among the most standard there could be. – ΑΘΩ Sep 21 '20 at 06:52
  • https://math.stackexchange.com/questions/292251/limit-of-s-n-int-limits-01-fracnxn-11x-dx-as-n-to-infty – Riemann Sep 21 '20 at 09:02

6 Answers6

9

We have: \begin{align} n\left (\frac{1}{n} - \frac{1}{n+1} + \cdots \right ) &= n \left ( \frac{1}{n(n+1)} + \frac{1}{(n+2)(n+3)} + \cdots \right ) \\ &\le n \left ( \frac{1}{n^2} + \frac{1}{(n+2)^2} + \cdots \right ) \\ &\le n \int_{n-2}^{\infty} \frac{1}{2x^2}dx = \frac{n}{2(n-2)} \end{align} Similarly, \begin{align} n\left (\frac{1}{n} - \frac{1}{n+1} + \cdots \right ) \ge \frac{n}{2(n+1)} \end{align}

Thus, by letting $n$ tends to infinity, we obtain \begin{align} \lim_{n\to \infty} {n \left ( \frac{1}{n} - \frac{1}{n+1} + \cdots \right )} = \frac{1}{2} \end{align}

sansae
  • 658
2

We have that

$$H_N=\sum_{k=1}^{N} \frac{1}k=\ln N+\gamma+\frac1{2N}+O\left(\frac1{N^2}\right)$$

then

$$\sum_{k=1}^{2N} \frac{(-1)^{k+1}}k=H_{2N}-H_{N}=\log {2}-\frac1{2N}+O\left(\frac1{N^2}\right)$$

and

$$\sum_{k=n}^{2N} \frac{(-1)^{k+1}}k=\sum_{k=1}^{2N} \frac{(-1)^{k+1}}k-\sum_{k=1}^{n-1} \frac{(-1)^{k+1}}k=$$

$$=-\frac1{2N}+O\left(\frac1{N^2}\right)+\frac1{2(n-1)}+O\left(\frac1{n^2}\right) \sim \frac1{2(n-1)}+O\left(\frac1{n^2}\right)$$

then

$$n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \ldots \right )\sim \frac n{2(n-1)}+O\left(\frac1{n}\right) \to \frac12$$

user
  • 154,566
1

Like @sansae $$L= \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right )$$ $$\implies L=\lim_{n \to \infty}n\left(\frac{1}{n(n+1)}+\frac{1}{(n+2)(n+3)}+\frac{1}{(n+4)(n+5)}+...+\frac{1}{(n+k)(n+k+1)}+...+\right)$$ But conver the limit to integral as $$\implies L= \lim_{n \to \infty}\frac{1}{n} \sum_{k=0}^{n} \frac{1}{(1+k/n)(1+(k+1)/n)}= \int_{0}^{1} (1+x)^{-2} dx=\frac{1}{2}.$$

Z Ahmed
  • 43,235
1

This would be my "napkin" heuristic:

Since $\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right)$ is the absolute value of the tail of a convergent series, it tends to zero. Therefore,

$$\begin{align*}\limsup_{n\to\infty} n&\left(\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \cdots\right) \\ = 1 &- \liminf_{n\to\infty}\, (n+1)\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right) \\ &+\lim_{n\to\infty}\left(\frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots\right) \\ = 1 &- \liminf_{n\to\infty} n\left(\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \cdots\right)\end{align*}$$

From here, we find that if the limit in question exists, it must equal $\frac{1}{2}.$

Brian Moehring
  • 21,088
0

As an alternative

$$n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \ldots \right )=$$

$$=n\left(\frac12 \frac1n+\frac12 \frac1n- \frac{1}{n+1} + \frac12\frac{1}{n+2}+\frac12\frac{1}{n+2}-\frac{1}{n+3}+\frac12\frac{1}{n+4}+\ldots\right)=$$

$$=\frac12+n\sum_{k=0}^\infty \frac{1}{(n+2k)(n+2k+1)(n+2k+2)} \to \frac12$$

indeed

$$n\sum_{k=0}^\infty \frac{1}{(n+2k)(n+2k+1)(n+2k+2)} \le n\sum_{k=0}^\infty \frac{1}{(n+2k)^3} =$$

$$=\frac1n\int_0^\infty \frac1{(1+2x)^3}dx=\frac 1{4n} \to 0$$

user
  • 154,566
0

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[5px,#ffd]{\lim_{n \to \infty} \int_{0}^{1}{nx^{n - 1} \over 1 + x}\,\dd x = {1 \over 2}}: \ {\Large ?}}$.

The integral can be evaluated, in the $\ds{n \to \infty}$-limit by means of the Laplace Method. Note that the "main contribution" to the integral comes from values of $\ds{x \lesssim 1}$ such that we make the change $\ds{x \mapsto 1 - x}$ to enforce the "main contribution" around $\ds{x \gtrsim 0}$. Namely, \begin{align} &\bbox[5px,#ffd]{\lim_{n \to \infty} \int_{0}^{1}{nx^{n - 1} \over 1 + x}\,\dd x} = \lim_{n \to \infty}\bracks{% n\int_{0}^{1}{\pars{1 - x}^{n - 1} \over 1 + \pars{1 - x}}\,\dd x} \\[5mm] = & \lim_{n \to \infty}\bracks{% n\int_{0}^{1}{\expo{\pars{n-1}\ln\pars{1 - x}} \over 2 - x}\,\dd x} = \lim_{n \to \infty}\bracks{% n\int_{0}^{\infty}{\expo{-\pars{n-1}x} \over 2 - 0}\,\dd x} \\[5mm] = &\ {1 \over 2}\lim_{n \to \infty}{n \over n - 1} = \bbx{\large{1 \over 2}} \\ & \end{align}

Felix Marin
  • 89,464