Question

Is the given series $\sum\limits_{1}^{\infty }{\dfrac{1}{2n+1}}$ convergent or divergent.

Answer 1

Now to check the nature of the series we will use a Limit comparison test. For this test we will compare the given series with $\sum{\dfrac{1}{n}}$ . Now we will find the value of $\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{2n+1}}{\dfrac{1}{n}} \right)$ . Using this we know that if the limit is positive and finite then the series both have the same nature. Now we know that the series $\sum{\dfrac{1}{n}}$ is divergent. Hence we can easily find the nature of the given series.

Complete answer:
Now consider the given series $\sum\limits_{1}^{\infty }{\dfrac{1}{2n+1}}$ .
Now to solve this problem we will use a limit comparison test.
Let us first understand the limit comparison test.
Let us say we have two series $\sum{{{A}_{n}}}$ and $\sum{{{B}_{n}}}$ .
Then we take the ratio $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{A}_{n}}}{{{B}_{n}}}$
Now if the ratio is finite and positive then the series have same nature
Which means either both series converge or both series diverge.
Consider the series $\sum{\dfrac{1}{n}}$ and $\sum{\dfrac{1}{2n}}$ .
Now let us take the ratio $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{1}{2n}}{\dfrac{1}{n}}=\dfrac{1}{2}$
Now we have that the ratio $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{\dfrac{1}{2n}}{\dfrac{1}{n}}=\dfrac{1}{2}$ is positive and finite.
Hence we can say that either both series are convergent or both series are divergent.
But we know that the series $\sum{\dfrac{1}{n}}$ is divergent.
Hence by limit comparison test we can say that the series $\sum{\dfrac{1}{2n}}$ is also divergent.
Now consider the given series $\sum\limits_{1}^{\infty }{\dfrac{1}{2n+1}}$
Now let us compare the given series with the series $\sum\limits_{1}^{\infty }{\dfrac{1}{n}}$
Consider $\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{\dfrac{1}{2n+1}}{\dfrac{1}{n}} \right)$
$\Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{n}{2n+1} \right)$
Now dividing the numerator and denominator by n we get,
$\begin{aligned} & \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{n}{2n+1} \right)=\underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2+\dfrac{1}{n}} \right) \\\ & \Rightarrow \underset{n\to \infty }{\mathop{\lim }}\,\left( \dfrac{n}{2n+1} \right)=\dfrac{1}{2+0}=\dfrac{1}{2} \\\ \end{aligned}$
Since the limit is finite and positive we can say that the series are of the same nature which means either both the series are convergent or both the series are divergent.
But we know that the series $\sum\limits_{1}^{\infty }{\dfrac{1}{n}}$ is divergent.
Hence, with the help of a limit comparison test we can say that the series $\sum\limits_{1}^{\infty }{\dfrac{1}{2n+1}}$ is also divergent.
Hence the given series is divergent.

Note: Now in limit comparison test note that if the ratio $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{A}_{n}}}{{{B}_{n}}}$ is finite and positive we have either both series converge or both diverge. If the ratio is zero and if $\sum{{{B}_{n}}}$ converges then $\sum{{{A}_{n}}}$ also converges. But if $\sum{{{B}_{n}}}$ diverges then we cannot say anything about $\sum{{{A}_{n}}}$ . Also if the ratio is infinity then if $\sum{{{B}_{n}}}$ diverges then $\sum{{{A}_{n}}}$ also diverges. But if $\sum{{{B}_{n}}}$ converges then we cannot say anything about $\sum{{{A}_{n}}}$ .