Series evaluation
- Tolaso J Kos
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Series evaluation
Taking for granted that: $\displaystyle \tan \frac{x}{2}=\cot \frac{x}{2}-2\cot x$ evaluate the series:
$$\sum_{n=1}^{\infty }\frac{1}{2^n}\tan \frac{x}{2^n}$$
$$\sum_{n=1}^{\infty }\frac{1}{2^n}\tan \frac{x}{2^n}$$
Imagination is much more important than knowledge.
- Grigorios Kostakos
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Re: Series evaluation
\(\color{gray}\bullet\) For \(x=0\) the series is equal to \(0\).
\(\color{gray}\bullet\) For \(x\neq 0\) we have that \begin{align*}
S_n(x)&=\mathop{\sum}\limits_{k=1}^{n}\frac{1}{2^k}\tan \bigl({\tfrac{x}{2^k}}\bigr)\\
&=\mathop{\sum}\limits_{k=1}^{n}\Bigl({\frac{1}{2^k}\cot \bigl({\tfrac{x}{2^k}}\bigr)-\frac{1}{2^{k-1}}\cot \bigl({\tfrac{x}{2^{k-1}}}\bigr)}\Bigr)\\
&=\mathop{\sum}\limits_{k=1}^{n}\frac{1}{2^k}\cot\bigl({\tfrac{x}{2^k}}\bigr)-\mathop{\sum}\limits_{k=1}^{n}\frac{1}{2^{k-1}}\cot\bigl({\tfrac{x}{2^{k-1}}}\bigr)\\
&=\mathop{\sum}\limits_{k=1}^{n}\frac{1}{2^k}\cot\bigl({\tfrac{x}{2^k}}\bigr)-\mathop{\sum}\limits_{k=0}^{n-1}\frac{1}{2^{k}}\cot\bigl({\tfrac{x}{2^{k}}}\bigr)\\
&=\frac{1}{2^n}\cot\bigl({\tfrac{x}{2^n}}\bigr)-\cot{x}\,.
\end{align*} Because \[\mathop{\lim}\limits_{n\to \infty}\frac{1}{2^n}\cot\bigl({\tfrac{x}{2^n}}\bigr)=\frac{1}{x}\,,\] we have that \[\sum_{n=1}^{\infty }\frac{1}{2^n}\tan \bigl({\tfrac{x}{2^n}}\bigr)=\mathop{\lim}\limits_{n\to \infty}S_n(x)=\frac{1}{x}-\cot{x}\,.\]
\(\color{gray}\bullet\) For \(x\neq 0\) we have that \begin{align*}
S_n(x)&=\mathop{\sum}\limits_{k=1}^{n}\frac{1}{2^k}\tan \bigl({\tfrac{x}{2^k}}\bigr)\\
&=\mathop{\sum}\limits_{k=1}^{n}\Bigl({\frac{1}{2^k}\cot \bigl({\tfrac{x}{2^k}}\bigr)-\frac{1}{2^{k-1}}\cot \bigl({\tfrac{x}{2^{k-1}}}\bigr)}\Bigr)\\
&=\mathop{\sum}\limits_{k=1}^{n}\frac{1}{2^k}\cot\bigl({\tfrac{x}{2^k}}\bigr)-\mathop{\sum}\limits_{k=1}^{n}\frac{1}{2^{k-1}}\cot\bigl({\tfrac{x}{2^{k-1}}}\bigr)\\
&=\mathop{\sum}\limits_{k=1}^{n}\frac{1}{2^k}\cot\bigl({\tfrac{x}{2^k}}\bigr)-\mathop{\sum}\limits_{k=0}^{n-1}\frac{1}{2^{k}}\cot\bigl({\tfrac{x}{2^{k}}}\bigr)\\
&=\frac{1}{2^n}\cot\bigl({\tfrac{x}{2^n}}\bigr)-\cot{x}\,.
\end{align*} Because \[\mathop{\lim}\limits_{n\to \infty}\frac{1}{2^n}\cot\bigl({\tfrac{x}{2^n}}\bigr)=\frac{1}{x}\,,\] we have that \[\sum_{n=1}^{\infty }\frac{1}{2^n}\tan \bigl({\tfrac{x}{2^n}}\bigr)=\mathop{\lim}\limits_{n\to \infty}S_n(x)=\frac{1}{x}-\cot{x}\,.\]
Grigorios Kostakos
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