Can we find $ \lim_{n \to \infty } n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + ... \right ) $? 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?
 A: 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}
A: 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
$$n\left ( \frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+2} - \frac{1}{n+3} + \ldots \right )\sim \frac n{2(n-1)}+O\left(\frac1{n}\right) \to \frac12$$
A: 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
$$\implies L= \lim_{n \to \infty}\frac{1}{n} \sum_{k=0}^{n} \frac{1}{(1+k/n)(1+(k+1)/n)}=
\int_{0}^{1} (1+x)^{-2} dx=\frac{1}{2}.$$
A: 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 $\frac{1}{2}.$
A: 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
$$n\sum_{k=0}^\infty \frac{1}{(n+2k)(n+2k+1)(n+2k+2)} \le n\sum_{k=0}^\infty \frac{1}{(n+2k)^3} =$$
$$=\frac1n\int_0^\infty \frac1{(1+2x)^3}dx=\frac 1{4n} \to 0$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\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}
