lim ∑r/1.3.5.7.9....2r+1 (limit from r=1 to n)is equal to

n tends to infinity.

To evaluate the limit of the series given by the expression \(\lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{1 \cdot 3 \cdot 5 \cdots (2r - 1)}\), we need to analyze the behavior of the terms in the sum as \(n\) approaches infinity. This series involves the product of odd numbers in the denominator, which can be expressed in a more manageable form.

Understanding the Denominator

The denominator \(1 \cdot 3 \cdot 5 \cdots (2r - 1)\) represents the product of the first \(r\) odd numbers. This product can be expressed using factorials. Specifically, it can be shown that:

• The product of the first \(r\) odd numbers is equal to \(\frac{(2r)!}{2^r \cdot r!}\).

Thus, we can rewrite the term in the sum as follows:

Rewriting the Series

Substituting this into our original expression gives:

\(\sum_{r=1}^{n} \frac{r}{1 \cdot 3 \cdot 5 \cdots (2r - 1)} = \sum_{r=1}^{n} \frac{r \cdot 2^r \cdot r!}{(2r)!}\)

Analyzing the Limit

Now, we need to analyze the behavior of the terms as \(n\) approaches infinity. The term \(\frac{r \cdot 2^r \cdot r!}{(2r)!}\) can be simplified further. Using Stirling's approximation, which states that \(n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\), we can approximate the factorials in the denominator:

\((2r)! \sim \sqrt{4 \pi r} \left(\frac{2r}{e}\right)^{2r}\)

Finding the Asymptotic Behavior

Thus, we can express our term as:

\(\frac{r \cdot 2^r \cdot r!}{(2r)!} \sim \frac{r \cdot 2^r \cdot \sqrt{2 \pi r} \left(\frac{r}{e}\right)^r}{\sqrt{4 \pi r} \left(\frac{2r}{e}\right)^{2r}} = \frac{r \cdot 2^r \cdot \sqrt{2 \pi r} \cdot r^r \cdot e^{2r}}{\sqrt{4 \pi r} \cdot 2^{2r} \cdot r^{2r}} = \frac{e^{2r} \cdot r}{\sqrt{2 \pi r} \cdot 2^{r} \cdot r^{r}} \cdot \frac{1}{\sqrt{4 \pi r}}\)

As \(r\) becomes large, the term \(\frac{r}{(2r)!}\) decreases rapidly, leading to the conclusion that the series converges.

Final Result

As \(n\) approaches infinity, the sum converges to a finite limit. In fact, it can be shown that:

\(\lim_{n \to \infty} \sum_{r=1}^{n} \frac{r}{1 \cdot 3 \cdot 5 \cdots (2r - 1)} = 1\)

This means that the series converges to 1 as \(n\) tends to infinity. Therefore, the limit of the given series is:

1