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

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!

Robert Z
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fiksx
  • 1,059

3 Answers3

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, $$2\sum_{k=0}^\infty\dfrac{(-1)^{k+1}}{2k+1}=\dfrac\pi2$$

lab bhattacharjee
  • 274,582
  • See also : https://math.stackexchange.com/questions/594838/taylor-series-for-log1x-and-its-convergence – lab bhattacharjee Mar 21 '18 at 16:34
  • thanks!!! but I don't understand where the $i$ came from? and why do you need to use $ln$? – fiksx Mar 21 '18 at 16:54
  • hello, I tried to understand your answer, but still have some few questions, why did you use $ ln(1+ x)$? and also how can you change it to $\implies\ln(1+i)-\ln(1-i)=2\sum_{k=0}^\infty\dfrac{i^{2k+1}}{2k+1}$ thanks!? – fiksx Mar 23 '18 at 02:07
0

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 $$1-\arctan(1)=1-\frac{\pi}{4}.$$

Robert Z
  • 145,942
  • thanks I got this!. $ arctan(1)=1-\frac{1}{3}+\frac{1}{5}..= 1-arctan(1)=1-\pi/4$ , but that Riemann sum is totally wrong(?) because of alternating(?) – fiksx Mar 21 '18 at 17:12
  • No, because $\frac{(-1)^{k+1}}{x}$ is not integrable over $(0,1]$ – Robert Z Mar 21 '18 at 17:13
  • @vik: actually you may remove the alternating signs by coupling adjacent terms, such that the problem boils down to computing $$\sum_{j\geq 1}\frac{1}{16j^2-1},$$ but Riemann sums are not the most suited tool for the explicit computation of such series. – Jack D'Aurizio Mar 21 '18 at 19:06
  • @RobertZ thankyou so much! , is it because $\frac{1}{x} $ diverge? it produce $\infty$ – fiksx Mar 23 '18 at 02:17
  • @vik Yes, that's right. – Robert Z Mar 23 '18 at 05:17
0

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Felix Marin
  • 89,464
  • thankyou I try to understand, is this using complex plane? and also how can you change it to $\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}}} \$? – fiksx Mar 23 '18 at 01:58
  • @vik Note that $-1 = \mathrm{i}^{2}$ and "sum over $\texttt{odd}$" is equat to "sum over $\texttt{all}$" minus "sum over $\texttt{even}$". I put $\Re$ at the very begining because I know the final result will be a $\texttt{real}$ number. It makes easier the whole evaluation. – Felix Marin Mar 23 '18 at 16:10
  • thankyou!! but still so hard for me to understand it, and I didn't understand how you get $-\Im ({ \lim_{n \to \infty}}\sum_{k = 3}^{2n + 1}{i^{k} \over k}-\sum_{k = 2}^{n}{i^{2k} \over 2k})$ , can you explain more how you got the "sum" $\sum_{k = 3}^{2n + 1}{i^{k} \over k}$ and why summation index start at $3$ and end at $2n-1$? and why you changed from $\Re$ to $-\Im$? – fiksx Mar 24 '18 at 07:35
  • @vik The original index $2k + 1$ goes from $3 = 2 \times 1 + 1$ to $2n +1$. As the sum is over the $\color{red}{\texttt{odd}}$ values, it's equivalent to the sum over $\color{red}{\texttt{all}}$ allowed values (from $3$ to $2n + 1$) minus the $\color{red}{\texttt{even}}$ values (from $3$ to $2n + 1$, the even values are $4,6,\ldots,2n$ such that indexes in $k = 2k/2$ goes from $2 = 4/2$ to $n = 2n/2$). – Felix Marin Mar 24 '18 at 17:22
  • ok i got it! but $\Im ({\lim_{n \to \infty}{{\color{red}{i}+ {\frac{\color{red}{i^2}}{2}} + \sum_{k = 1}^{2n + 1}{i^{k} \over k}} - {1 \over 2}\sum_{k = 2}^{n}{\pars{-1}^{k} \over k})}} $ if it start from $k=1$ like in red, why there is negative on your post? and also why you have to change index to 1? and also limit $n-> \infty$ here will make summation become zero? and how you got $\Im ln(1-i)$? thankyou!! sorry im asking too many questions ! – fiksx Mar 25 '18 at 17:18
  • @vik I have to "restart" the sum from $k = 1$ because, in that case, I'll have the $\ln$ series. – Felix Marin Mar 25 '18 at 22:02