Determine whether the series is convergent or divergent.
\[
1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots
\]

Step 1 ：We are given the series \(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots\). This is a harmonic series with odd denominators. The general term of the series is \(\frac{1}{2n-1}\) where \(n\) is the term number. The harmonic series is known to be divergent. However, this series only includes the terms of the harmonic series with odd denominators, so we need to check if this series is convergent or divergent.

Step 2 ：We can use the integral test to determine the convergence of the series. The integral test states that if \(f(x)\) is a positive, continuous, and decreasing function on the interval \([1, \infty)\), then the infinite series \(\sum_{n=1}^{\infty} f(n)\) is convergent if and only if the improper integral \(\int_{1}^{\infty} f(x) dx\) is convergent.

Step 3 ：In this case, our function \(f(x)\) is \(\frac{1}{2x-1}\). We can integrate this function from 1 to infinity to see if the integral converges or diverges.

Step 4 ：The integral of the function \(\frac{1}{2x-1}\) from 1 to infinity is infinity, which means the integral diverges. According to the integral test, this means that the original series also diverges.

Step 5 ：Final Answer: The series \(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots\) is \(\boxed{\text{divergent}}\).