# Find the value of: <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD">

Find the value of:
$\sum _{k=1}^{\mathrm{\infty }}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({2}^{k+1}+\frac{1}{{2}^{k}}\right)$
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lorienoldf7
You know
$\mathrm{cot}\left(a-b\right)=\frac{\mathrm{cot}a.\mathrm{cot}b+1}{\mathrm{cot}b-\mathrm{cot}a}$
so
$\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(\frac{x.y+1}{y-x}\right)=\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(y\right)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(x\right)$
$\sum _{k=1}^{\mathrm{\infty }}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({2}^{k+1}+\frac{1}{{2}^{k}}\right)=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty }}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(\frac{{2}^{k+1}{.2}^{k}+1}{{2}^{k}}\right)=\phantom{\rule{0ex}{0ex}}$
you can rewrite ${2}^{k}$ as ${2}^{k+1}-{2}^{k}$ so you will have
$\sum _{k=1}^{\mathrm{\infty }}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(\frac{{2}^{k+1}{.2}^{k}+1}{{2}^{k+1}-{2}^{k}}\right)=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty }}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({2}^{k}\right)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left({2}^{k+1}\right)=\phantom{\rule{0ex}{0ex}}\sum _{k=1}^{\mathrm{\infty }}f\left(k\right)-f\left(k+1\right)=\phantom{\rule{0ex}{0ex}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(2\right)-\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(\mathrm{\infty }\right)=\phantom{\rule{0ex}{0ex}}\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(2\right)-0$
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Sattelhofsk
HINT:
See Are $\mathrm{a}\mathrm{r}\mathrm{c}\mathrm{c}\mathrm{o}\mathrm{t}\left(x\right)$ and $\mathrm{arctan}\left(1/x\right)$ the same function?
Can you use
$\frac{{2}^{k}}{{2}^{2k+1}+1}=\frac{{2}^{k+1}-{2}^{k}}{1+{2}^{k}\cdot {2}^{k+1}}$
to find
$\mathrm{arctan}\frac{{2}^{k}}{{2}^{2k+1}+1}=\mathrm{arctan}\left({2}^{k+1}\right)-\mathrm{arctan}\left({2}^{k}\right)$