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QuestionFind the limit: lim⁡n→∞(1+x)(1+x2)(1+x4)⋯(1+x2n)\lim _{n \rightarrow \infty}(1+x)(1+x^{2})(1+x^{4}) \cdots(1+x^{2^{n}}) for ∣x∣<1|x|<1.

Studdy Solution

STEP 1

Assumptions1. We are given a limit as n approaches infinity. . The function is a product of terms of the form (1+xk)(1+x^{^{k}}) where k ranges from0 to n.
3. The absolute value of x is less than1.

STEP 2

We know that for ∣x∣<1|x|<1, the geometric series formula can be used∑k=0∞xk=11−x\sum_{k=0}^{\infty} x^{k} = \frac{1}{1-x}

STEP 3

Now, we can rewrite each term of the form (1+x2k)(1+x^{2^{k}}) as a geometric series(1+x2k)=∑j=0∞x2kj(1+x^{2^{k}}) = \sum_{j=0}^{\infty} x^{2^{k}j}

STEP 4

The given function can now be written as the product of these infinite geometric series∏k=0n∑j=0∞x2kj\prod_{k=0}^{n} \sum_{j=0}^{\infty} x^{2^{k}j}

STEP 5

Next, we can rewrite the product of sums as a single sum over all sequences of nonnegative integers (j0,j1,...,jn)(j0, j1, ..., j_n)∑j0=0∞∑j1=0∞⋯∑jn=0∞x20j0+21j1+...+2njn\sum_{j0=0}^{\infty} \sum_{j1=0}^{\infty} \cdots \sum_{j_n=0}^{\infty} x^{2^{0}j0 +2^{1}j1 + ... +2^{n}j_n}

STEP 6

We can then rewrite the exponent as a single integer m∑m=0∞amxm\sum_{m=0}^{\infty} a_m x^{m}where ama_m is the number of ways to write mm as a sum of distinct powers of2.

STEP 7

We know that every nonnegative integer can be written uniquely as a sum of distinct powers of2 (this is the binary representation of integers). Therefore, am=1a_m =1 for all mm.

STEP 8

So, the function simplifies to∑m=0∞xm\sum_{m=0}^{\infty} x^{m}

STEP 9

Finally, we can use the geometric series formula to find the limit as n approaches infinitylim⁡n→∞∑m=∞xm=lim⁡n→∞−x\lim_{n \rightarrow \infty} \sum_{m=}^{\infty} x^{m} = \lim_{n \rightarrow \infty} \frac{}{-x}

STEP 10

Since the limit does not actually depend on n, we can drop the limit−x\frac{}{-x}So, the solution to the given limit problem is −x\frac{}{-x}, for ∣x∣<|x|<.

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