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**Hint:**We can solve this question by drawing the graph of $\dfrac{1}{2x+1}$ . We can prove some series is divergent if some less value than the series is divergent. We can prove some series convergent if some greater value than the series is convergent.

**Complete step by step answer:**

In the series $\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}.......$ we can see the denominators are 3,5,7….

We can write the series $\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}.......=\sum\limits_{n=1}^{\infty }{\dfrac{1}{2n+1}}$

Let’s the graph of $\dfrac{1}{2x+1}$

Lets mark the area under the curve when $1\le x<\infty $

The region with deep blue color is our required area we can calculate the area by integrating

$\dfrac{1}{2x+1}$

From 1 to infinity

So the area under the curve =$\int\limits_{1}^{\infty }{\dfrac{1}{2x+1}dx}$

$=\left[ \dfrac{1}{2}\ln \left( 2x+1 \right) \right]_{1}^{\infty }$

We know that logarithm is tend to infinity when x tends to infinity

So the area under the curve tends to infinity.

Now let’s look at the series given in the question $\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}.......$

If can write the series as

$\dfrac{1}{3}\times 1+\dfrac{1}{5}\times 1+\dfrac{1}{7}\times 1..........$

We represent the series in as area in a Cartesian plane and compare with above area under the curve

_{}

All the rectangle represents the area

$\dfrac{1}{3}\times 1+\dfrac{1}{5}\times 1+\dfrac{1}{7}\times 1..........$

We can clearly see the sum of the area of all rectangles is greater than the area under the curve from 1 to infinity.

If the area under the curve tends to infinity then the area of all rectangles will automatically tend to infinity.

So the sum of the series

$\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}+\dfrac{1}{11}.......$

Will tend to infinity. So the series is divergent.

**Note:**

If nth term of a series is a nonzero number when n tends to infinity then the series is definitely divergent but the reverse is not true that means if nth term tends to 0 when n tends to infinity that does not mean the series will be convergent like we have seen in the above example. In the above series the nth term tends to 0 when n tends to infinity but the series

Is divergent.

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