Find $\lim_{n \to \infty}\sum_{k=1}^n \frac{(-1)^{k+1}}{2k+1}$

Question

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

How can I find limit value of this? I tried using Riemann, by supposing $x_k=\frac{2k+1}{n}$.

$\lim_{n \to \infty}\sum_{k=1}^n \frac{(-1)^{k+1}}{ \frac{2k+1}{n}} \frac{1}{n}$
$x_{k+1}-x_k=\frac{2}{n}$
$\frac{1}{2} \lim_{n \to \infty}\sum_{k=1}^n \frac{(-1)^{k+1}}{ \frac{2k+1}{n}} \frac{1}{n}$
$\frac{1}{2} \lim_{n \to \infty}\int_0^1\frac{(-1)^{k+1}}{x} dx$
is this possible? thankyou!

Answer 1

$$\dfrac{(-1)^{k+1}}{2k+1}=i\cdot\dfrac{i^{2k+1}}{2k+1}$$

Now $\ln(1+x)=x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots$

$$\implies\ln(1+i)-\ln(1-i)=2\sum_{k=0}^\infty\dfrac{i^{2k+1}}{2k+1}$$

Now $\ln(1+i)-\ln(1-i)\equiv\ln\dfrac{1+i}{1-i}\pmod{2\pi i}$

$\equiv\ln\dfrac{(1+i)^2}2\equiv\ln(i)\equiv i\dfrac\pi2$

So, equating the real parts, $$2\sum_{k=0}^\infty\dfrac{(-1)^{k+1}}{2k+1}=\dfrac\pi2$$

Answer 2

I don't think that Riemann sum is the right tool. Use the Taylor series of $x-\arctan(x)$ for $x=1$ centered at $0$. See Why is $\arctan(x)=x-x^3/3+x^5/5-x^7/7+\dots$? Hence the given sum is equal to $$1-\arctan(1)=1-\frac{\pi}{4}.$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \lim_{n \to \infty}\sum_{k = 1}^{n}{\pars{-1}^{k + 1} \over 2k + 1} & = \Re\bracks{\ic\,\lim_{n \to \infty}\sum_{k = 1}^{n}{\ic^{2k + 1} \over 2k + 1}} = -\Im\bracks{\lim_{n \to \infty}\pars{% \sum_{k = 3}^{2n + 1}{\ic^{k} \over k} - \sum_{k = 2}^{n}{\ic^{2k} \over 2k}}} \\[5mm] & = -\Im\braces{\lim_{n \to \infty}\bracks{% \pars{-\ic + {1 \over 2} + \sum_{k = 1}^{2n + 1}{\ic^{k} \over k}} - {1 \over 2}\sum_{k = 2}^{n}{\pars{-1}^{k} \over k}}} \\[5mm] & = 1 + \Im\ln\pars{1 - \ic} = \bbx{1 - {\pi \over 4}} \approx 0.2146 \end{align}