Arccot series summation

Question

Find the value of:

$$\sum_{k=1}^{\infty}\mathrm{arccot}\left(2^{k+1}+\dfrac{1}{2^k}\right)$$

I think we need to use telescoping method, but I am not able to convert this.

Answer 1

You know $$\cot(a-b)=\dfrac{\cot a.\cot b+1}{\cot b-\cot a}$$so $$\mathrm{arccot}(\dfrac{x.y+1}{y-x})=\mathrm{arccot}(y)-\mathrm{arccot}(x)$$ $$\sum_{k=1}^{\infty}\mathrm{arccot}\left(2^{k+1}+\dfrac{1}{2^k}\right)=\\ \sum_{k=1}^{\infty}\mathrm{arccot}\left(\dfrac{2^{k+1}.2^{k}+1}{2^k}\right)=\\$$ you can rewrite $2^k$ as $2^{k+1}-2^k$ so you will have $$\sum_{k=1}^{\infty}\mathrm{arccot}\left(\dfrac{2^{k+1}.2^{k}+1}{2^{k+1}-2^{k}}\right)=\\ \sum_{k=1}^{\infty}\mathrm{arccot}\left(2^{k}\right)-\mathrm{arccot}\left(2^{k+1}\right)=\\ \sum_{k=1}^{\infty}f(k)-f(k+1)=\\\mathrm{arccot}(2)-\mathrm{arccot}(\infty)=\\\mathrm{arccot}(2)-0$$

Answer 2

HINT:

See Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?

Can you use $$\dfrac{2^k}{2^{2k+1}+1}=\dfrac{2^{k+1}-2^k}{1+2^k\cdot2^{k+1}}$$

to find $$\arctan\dfrac{2^k}{2^{2k+1}+1}=\arctan(2^{k+1})-\arctan(2^k)$$