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I got the result below during my research.

$$1=\frac{1}{1+a_1}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)}+... \tag 1$$

$$1=\frac{1}{1+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (1+a_j)} $$

$$1+a_1=1+\frac{a_1}{1+a_2}+\frac{a_1a_2}{(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+...$$

$$a_1=\frac{a_1}{1+a_2}+\frac{a_1a_2}{(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+...$$

We can get the same relation but without $a_1$ $$1=\frac{1}{1+a_2}+\frac{a_2}{(1+a_2)(1+a_3)}+\frac{a_2a_3}{(1+a_2)(1+a_3)(1+a_4)}+...$$

Examples:

Example-1: $a_n=c$

$$1=\frac{1}{1+c}+\frac{c}{(1+c)^2}+\frac{c^2}{(1+c)^3}+\frac{c^3}{(1+c)^4}+...$$

$$1+c=1+\frac{c}{1+c}+\frac{c^2}{(1+c)^2}+\frac{c^3}{(1+c)^3}+...$$

We know that if $\frac{c}{(1+c)}<1$

then $$1+\frac{c}{1+c}+\frac{c^2}{(1+c)^2}+\frac{c^3}{(1+c)^3}+...=\frac{1}{1-\frac{c}{(1+c)}}=1+c$$

Example-2: $a_n=x^{2^{n-1}}$

$$1=\frac{1}{1+x}+\frac{x}{(1+x)(1+x^2)}+ \frac{x^3}{(1+x)(1+x^2)(1+x^4)}+ \frac{x^7}{(1+x)(1+x^2)(1+x^4)(1+x^8)}+...$$

$$1=\frac{1-x}{1-x^2}+\frac{x(1-x)}{1-x^4}+\frac{x^3(1-x)}{1-x^8} +\frac{x^7(1-x)}{1-x^{16}} +...$$

$$1=\frac{1-x}{1-x^2}+\frac{x(1-x)}{1-x^4}+\frac{x^3(1-x)}{1-x^8} +\frac{x^7(1-x)}{1-x^{16}} +...$$

$$\frac{x}{1-x}=\frac{x}{1-x^2}+\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8} +\frac{x^8}{1-x^{16}} +...$$

If we put $x---->x^2$

$$\frac{x^2}{1-x^2}=\frac{x^2}{1-x^4}+\frac{x^4}{1-x^8} +\frac{x^8}{1-x^{16}} +....$$

$$\frac{x}{1-x}=\frac{x}{1-x^2}+\frac{x^2}{1-x^2}=\frac{x(1+x)}{1-x^2}$$

Example-3: $a_n=n$

$$1=\frac{1}{2}+\frac{1}{2.3}+\frac{1.2}{2.3.4}+\frac{1.2.3}{2.3.4.5}+... $$

$$1=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+... $$

$$\frac{1}{2}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+... $$

I wonder which general conditions are required for $a_n$ that Equation $ 1 $ is true.

Thanks for answers and comments.

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Looks like you need $\prod_{i=1}^k \frac{a_i}{1+a_i}$ to converge to zero as $k \rightarrow \infty$. –  polkjh May 24 '13 at 8:06
    
@polkjh I proved the relation and you are right about the condition for $a_n$. Thanks –  Mathlover May 24 '13 at 12:48
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1 Answer 1

I proved the relation and required condition for $a_n$. I would like to share it.

$$(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... =A\tag 1$$

$$A/A=1=\frac{(1+a_2)(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... }+a_1\frac{(1+a_2)(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 2$$

$$1=\frac{1}{(1+a_1)}+a_1\frac{(1+a_2)(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 3$$

$$1=\frac{1}{(1+a_1)}+a_1\frac{(1+a_3)(1+a_4)(1+a_5)...+a_2(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 4$$

$$1=\frac{1}{(1+a_1)}+\frac{a_1}{(1+a_1)(1+a_2)}+a_1a_2\frac{(1+a_3)(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 5$$

$$1=\frac{1}{(1+a_1)}+\frac{a_1}{(1+a_1)(1+a_2)}+a_1a_2\frac{(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... }+a_1a_2\frac{a_3(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 6$$

$$1=\frac{1}{(1+a_1)}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+a_1a_2a_3\frac{(1+a_4)(1+a_5)...}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... } \tag 7$$

If we continue in that way to create series then the last term will be the same as @polkjh wrote in comments.

$$\frac{a_1a_2a_3a_4a_5....}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)(1+a_5)... }\tag 8$$

The last term must go to zero otherwise the relation I wrote above must be as shown below

$$1-\frac{\prod\limits_{j=1}^{\infty}a_j}{\prod\limits_{j=1}^{\infty} (1+a_j)}=\frac{1}{1+a_1}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)}+... \tag 9$$

$$1-\frac{\prod\limits_{j=1}^{\infty}a_j}{\prod\limits_{j=1}^{\infty} (1+a_j)}=\frac{1}{1+a_1}+\sum\limits_{k=1}^ \infty\frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (1+a_j)} \tag {10}$$

If $\frac{\prod\limits_{j=1}^{\infty}a_j}{\prod\limits_{j=1}^{\infty} (1+a_j)}=0$ then the relation will be

$$1=\frac{1}{1+a_1}+\frac{a_1}{(1+a_1)(1+a_2)}+\frac{a_1a_2}{(1+a_1)(1+a_2)(1+a_3)}+\frac{a_1a_2a_3}{(1+a_1)(1+a_2)(1+a_3)(1+a_4)}+... \tag {11}$$

Also The formula can be written for n terms. $$1-\frac{\prod\limits_{j=1}^{n}a_j}{\prod\limits_{j=1}^{n} (1+a_j)}=\frac{1}{1+a_1}+\sum\limits_{k=1}^ {n-1} \frac{\prod\limits_{j=1}^{k}a_j}{\prod\limits_{j=1}^{k+1} (1+a_j)} \tag {12}$$

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(+1):Really interesting!Well done. –  shaswata Jun 1 '13 at 18:14
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