\(\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...\)

To determine: The convergence and divergence of the given series by Integral test.

Explanation:

Integral Test: Let f(x) be a positive, continuous and decreasing function for \(x\in[1,\infty)\) and f(n)=a_n where n is any natural number then the improper integral \(\int_1^\infty f(x)dx\) and infinite \(\sum_{n=1}^\infty a_n\) both converge or diverge simultaneously.

Here the infinite sum is \(\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...=\sum_{n=1}^\infty\frac{1}{2n+1}\)

Therefore \(a_n=\frac{1}{2n+1}\)

So let \(f(x)=\frac{1}{2x+1}\) for \(x\in[1,\infty)\)

Then clearly f(x) is continuous, positive, and decreasing for \(x\in[1,\infty)\) and \(f(n)=\frac{1}{2n+1}=a_n\) for any natural number n.

Hence Integral Test can be applied for \(f(x)=\frac{1}{2x+1}\) and \(a_n\frac{1}{2n+1}\)

Now checking for the convergence or divergence of the improper integral

\(\int_1^\infty f(x)dx\) for \(f(x)=\frac{1}{2x+1}\)

So,

\(\int_1^\infty f(x)dx\)

\(=\lim_{t\rightarrow\infty}\int_1^t\frac{1}{2x+1}dx\)

\(=\frac12\lim_{t\rightarrow\infty}\int_1^t\frac{2}{2x+1}dx\) [multiplying and dividing by 2]

\(=\frac12\lim_{t\rightarrow\infty}[\ln(2x+1)]_1^t\quad[\because\int\frac{f'(x)}{f(x)}dx=\ln[f(x)]]\)

\(=\frac12\lim_{t\rightarrow\infty}[\ln(2t+1)-\ln3]\)

\(=\infty\)

Hence \(\int_1^\infty f(x)dx=\int_1^\infty\frac{1}{2x+1}dx\) diverges

Therefore by Integral test

\(\sum_{n=1}^\infty\frac{1}{2n+1}=\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...\) also diverge.

Answer: The series \(\frac13+\frac15+\frac17+\frac19+\frac{1}{11}+...\) is divergent