Question

The limit $\lim _{n \rightarrow \infty}\left(1+\frac{1}{5}\right)\left(1+\frac{1}{5^{2}}\right)\left(1+\frac{1}{5^{4}}\right) \ldots\left(1+\frac{1}{5^{2^{*}}}\right)$ is equal to (A) 0 (B) \(5 / 4\) (C) \(4 / 5\) (D) \(1 / 5\)

Short answer

A. The product does not converge. B. The product converges to \(\frac{26}{25}\). C. The product converges to \(\frac{35}{25}\). D. The product converges to \(\frac{37}{25}\). Answer: We can only be certain that the product converges to a finite value. Therefore, option (A) is incorrect, but we cannot definitively determine the exact value of the limit among options (B), (C), and (D).

Step-by-step solution

Identify the general term of the infinite product

The infinite product given in the exercise is as follows: \(\prod_{k=0}^{\infty}\left(1+\frac{1}{5^{2^k}}\right)\) Here, the general term of the product is \((1+\frac{1}{5^{2^k}})\), where \(k\) goes from \(0\) to \(\infty\).

Apply the formula for the infinite product

Let's denote the given infinite product by \(P\), so we can write: \(P = \lim_{n \rightarrow \infty}\prod_{k=0}^{n}\left(1+\frac{1}{5^{2^k}}\right)\) Now, we will take the logarithm of this product to convert it into a sum. We have: \(\ln(P) = \ln\left(\lim_{n \rightarrow \infty}\prod_{k=0}^{n}\left(1+\frac{1}{5^{2^k}}\right)\right)\) Using the properties of logarithms, we can rewrite this as a sum: \(\ln(P) = \lim_{n \rightarrow \infty}\sum_{k=0}^{n}\ln\left(1+\frac{1}{5^{2^k}}\right)\)

Analyze the behavior of the sum as \(n\) goes to infinity

Notice that the logarithm of the infinite product converges as \(n\) goes to infinity since the terms inside the logarithm are always positive and decreasing. So, we can take the limit inside the sum: \(\ln(P) = \sum_{k=0}^{\infty}\ln\left(1+\frac{1}{5^{2^k}}\right)\)

Calculate the infinite sum

Now, let's focus on calculating the infinite sum. We have: \(\ln(P) = \ln\left(1+\frac{1}{5^{2^0}}\right) + \ln\left(1+\frac{1}{5^{2^1}}\right) + \ln\left(1+\frac{1}{5^{2^2}}\right) + \ldots\) Using the properties of logarithms, this can be combined into a single logarithm: \(\ln(P) = \ln\left(\left(1+\frac{1}{5^{2^0}}\right)\left(1+\frac{1}{5^{2^1}}\right)\left(1+\frac{1}{5^{2^2}}\right)\ldots\right)\) This expression is equivalent to the original infinite product that we set out to find: \(\ln(P) = \ln\left(\prod_{k=0}^{\infty}\left(1+\frac{1}{5^{2^k}}\right)\right)\) So, we have shown that the logarithm of the infinite product converges and the sum goes to a finite value.

Find the limit of the product

We now have a relationship between the limit of the product and the logarithm of the limit of the product: \(P = \prod_{k=0}^{\infty}\left(1+\frac{1}{5^{2^k}}\right)\) Thus, \(\lim_{n \rightarrow \infty}\left(1+\frac{1}{5}\right)\left(1+\frac{1}{5^{2}}\right)\left(1+\frac{1}{5^{4}}\right)\ldots\left(1+\frac{1}{5^{2^{*}}}\right) = P\) To find the value of \(P\), we can calculate the exponential of the logarithm we have found: \(P = \exp(\ln(P))\) Since \(\ln(P)\) is a finite value, we can conclude that the infinite product converges to a finite value. Thus, we eliminate option (A) from the given choices. However, we don't have the necessary information to determine the exact value of the limit. The remaining options (B), (C), and (D) involve fractions with different numerators and denominators, so we cannot definitively determine the value of the limit.