$\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}({2}^{k+1}+{\displaystyle \frac{1}{{2}^{k}}})$

George Bray
2022-06-25
Answered

Find the value of:

$\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}({2}^{k+1}+{\displaystyle \frac{1}{{2}^{k}}})$

$\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}({2}^{k+1}+{\displaystyle \frac{1}{{2}^{k}}})$

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lorienoldf7

Answered 2022-06-26
Author has **19** answers

You know

$\mathrm{cot}(a-b)={\displaystyle \frac{\mathrm{cot}a.\mathrm{cot}b+1}{\mathrm{cot}b-\mathrm{cot}a}}$

so

$\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}({\displaystyle \frac{x.y+1}{y-x}})=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(y)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(x)$

$\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}({2}^{k+1}+{\displaystyle \frac{1}{{2}^{k}}})=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({\displaystyle \frac{{2}^{k+1}{.2}^{k}+1}{{2}^{k}}}\right)=\phantom{\rule{0ex}{0ex}}$

you can rewrite ${2}^{k}$ as ${2}^{k+1}-{2}^{k}$ so you will have

$\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({\displaystyle \frac{{2}^{k+1}{.2}^{k}+1}{{2}^{k+1}-{2}^{k}}}\right)=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({2}^{k}\right)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({2}^{k+1}\right)=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty}}f(k)-f(k+1)=\phantom{\rule{0ex}{0ex}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(2)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(\mathrm{\infty})=\phantom{\rule{0ex}{0ex}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(2)-0$

$\mathrm{cot}(a-b)={\displaystyle \frac{\mathrm{cot}a.\mathrm{cot}b+1}{\mathrm{cot}b-\mathrm{cot}a}}$

so

$\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}({\displaystyle \frac{x.y+1}{y-x}})=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(y)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(x)$

$\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}({2}^{k+1}+{\displaystyle \frac{1}{{2}^{k}}})=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({\displaystyle \frac{{2}^{k+1}{.2}^{k}+1}{{2}^{k}}}\right)=\phantom{\rule{0ex}{0ex}}$

you can rewrite ${2}^{k}$ as ${2}^{k+1}-{2}^{k}$ so you will have

$\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({\displaystyle \frac{{2}^{k+1}{.2}^{k}+1}{{2}^{k+1}-{2}^{k}}}\right)=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({2}^{k}\right)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({2}^{k+1}\right)=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty}}f(k)-f(k+1)=\phantom{\rule{0ex}{0ex}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(2)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(\mathrm{\infty})=\phantom{\rule{0ex}{0ex}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(2)-0$

Sattelhofsk

Answered 2022-06-27
Author has **5** answers

HINT:

See Are $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(x)$ and $\mathrm{arctan}(1/x)$ the same function?

Can you use

$\frac{{2}^{k}}{{2}^{2k+1}+1}}={\displaystyle \frac{{2}^{k+1}-{2}^{k}}{1+{2}^{k}\cdot {2}^{k+1}}$

to find

$\mathrm{arctan}{\displaystyle \frac{{2}^{k}}{{2}^{2k+1}+1}}=\mathrm{arctan}({2}^{k+1})-\mathrm{arctan}({2}^{k})$

See Are $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}(x)$ and $\mathrm{arctan}(1/x)$ the same function?

Can you use

$\frac{{2}^{k}}{{2}^{2k+1}+1}}={\displaystyle \frac{{2}^{k+1}-{2}^{k}}{1+{2}^{k}\cdot {2}^{k+1}}$

to find

$\mathrm{arctan}{\displaystyle \frac{{2}^{k}}{{2}^{2k+1}+1}}=\mathrm{arctan}({2}^{k+1})-\mathrm{arctan}({2}^{k})$

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