Question

If \(|x|<1\) then find the sum of the series \(\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{4 x^{3}}{1+x^{4}}+\frac{8 x^{7}}{1+x^{8}}+\ldots \ldots \infty\)

Short answer

Question: Find the sum of the infinite series: \(\frac{1}{1+x}+\frac{2x}{1+x^{2}}+\frac{4x^{3}}{1+x^{4}}+\frac{8x^{7}}{1+x^{8}}+\ldots\) Answer: The sum of the infinite series is given by \(S = \sum_{k=1}^{\infty} \frac{2^{k-1} x^{2^{k-1}}}{1 + x^{2^k}}\), valid for \(|x| < 1\).

Step-by-step solution

Recognize the pattern in the given series

To recognize the pattern, we can look at the numerators and denominators of the terms in the series. The numerators are powers of 2, and the denominators are in the form of \(1 + x^{2^n}\) where \(n\) is a non-negative integer. The terms can be written as: Term 1: \(\frac{1}{1+x} \rightarrow 2^{0}\cdot x^{0}\) Term 2: \(\frac{2x}{1+x^{2}} \rightarrow 2^{1}\cdot x^{2^{0}}\) Term 3: \(\frac{4x^{3}}{1+x^{4}} \rightarrow 2^{2}\cdot x^{2^{1}+2^{0}}\) Term 4: \(\frac{8x^{7}}{1+x^{8}} \rightarrow 2^{3}\cdot x^{2^{2}+2^{1}+2^{0}}\) And so on. So the general term can be written as: Term \(k\): \(T_k = \frac{2^{k-1} x^{2^{k-1} + 2^{k-2} + ... + 2^0}}{1 + x^{2^k}}\)

Use the geometric series formula to find the sum

Notice that each term \(T_k\) can be written as a geometric series: \(T_k = \frac{2^{k-1} x^{2^{k-1}}} {1 + x^{2^k}} \cdot \left(1 + x^{2^{k-1}} + x^{2\cdot2^{k-1}} + ... + x^{2^{k-1}(2^{k-1})} \right)\) Now, we can find the sum of the given series by adding all the terms, which is given by: \(S = \sum_{k=1}^{\infty} T_k\) Since each \(T_k\) is a geometric series with common ratio \(x^{2^{k-1}}\), we can find the sum using the geometric series formula: \(S = \sum_{k=1}^{\infty} \frac{2^{k-1} x^{2^{k-1}}} {1 + x^{2^k}} \cdot \frac{1 - (x^{2^{k-1}})^{2^{k-1}+1}}{1 - x^{2^{k-1}}}\). We will now simplify this expression to get the final answer.

Simplify the expression

Looking at the sum formula in Step 2, notice that the numerator in the geometric series formula \(1 - (x^{2^{k-1}})^{2^{k-1}+1}\) can be simplified as: \(1 - (x^{2^{k-1}})^{2^{k-1}+1} = 1 - x^{2^{k-1}(2^{k-1}+1)} = 1 - x^{2^k + 2^{k-1}}\) Now, we can substitute this back into the sum formula: \(S = \sum_{k=1}^{\infty} \frac{2^{k-1} x^{2^{k-1}} (1 - x^{2^k + 2^{k-1}})}{(1 + x^{2^k})(1 - x^{2^{k-1}})}\). Finally, we will factor the numerator to obtain a simpler sum expression: \(S = \sum_{k=1}^{\infty} \frac{2^{k-1} x^{2^{k-1}} (1 - x^{2^{k-1}})(1+x^{2^{k-1}})}{(1 + x^{2^k})(1 - x^{2^{k-1}})} = \sum_{k=1}^{\infty} \frac{2^{k-1} x^{2^{k-1}}}{1 + x^{2^k}}\) This is the final expression for the sum of the given series, valid for \(|x| < 1\).