Infinite Sum
- Tolaso J Kos
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Infinite Sum
Evaluate the sum : \( \displaystyle \frac{x}{x+1}+\frac{x^2}{\left ( x+1 \right )\left ( x^2+1 \right )}+\frac{x^4}{\left ( x+1 \right )\left ( x^2+1 \right )\left ( x^4+1 \right )}+\cdots \), whereas \(x>1 \).
Imagination is much more important than knowledge.
- Tolaso J Kos
- Administrator
- Posts: 867
- Joined: Sat Nov 07, 2015 6:12 pm
- Location: Larisa
- Contact:
Re: Infinite Sum
Finally a solution to this monster.
The sum telescopes (wow!!)
We note that: $$\begin{aligned}
\frac{x}{x+1} &=1-\frac{1}{x+1} \\
\frac{x^2}{\left ( 1+x \right )\left ( 1+x^2 \right )} &=\frac{1}{x+1}-\frac{1}{\left ( 1+x \right )\left ( 1+x^2 \right )} \\
\vdots &=\vdots \\
\frac{x^{2n}}{(1+x)(1+x^2)\cdots(1+x^{2n})}&= \frac{1}{(x+1)(1+x^2)\cdots(1+x^{2(n-1)})}- \\
&\;\;\;\;-\frac{1}{(1+x)(1+x^2)\cdots (1+x^{2n})}
\end{aligned}$$ Therefore the sum is \( 1 \).
The sum telescopes (wow!!)
We note that: $$\begin{aligned}
\frac{x}{x+1} &=1-\frac{1}{x+1} \\
\frac{x^2}{\left ( 1+x \right )\left ( 1+x^2 \right )} &=\frac{1}{x+1}-\frac{1}{\left ( 1+x \right )\left ( 1+x^2 \right )} \\
\vdots &=\vdots \\
\frac{x^{2n}}{(1+x)(1+x^2)\cdots(1+x^{2n})}&= \frac{1}{(x+1)(1+x^2)\cdots(1+x^{2(n-1)})}- \\
&\;\;\;\;-\frac{1}{(1+x)(1+x^2)\cdots (1+x^{2n})}
\end{aligned}$$ Therefore the sum is \( 1 \).
Imagination is much more important than knowledge.
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