Answer

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**Hint:**To solve this question we need to know the concept of ratio test which will be used to find the…… of the series. For finding whether the series is convergent or divergent, we will firstly find the value of ${{a}_{n+1}}$ and ${{a}_{n}}$. Then check that whether the value of $\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$ is less than, greater than or equal to $1$.

**Complete step by step answer:**

The question ask us to find whether the series given to us which is $1+\dfrac{x}{2}+\dfrac{{{x}^{2}}}{5}+\dfrac{{{x}^{3}}}{10}........\dfrac{{{x}^{n}}}{{{n}^{2}}+1}+.......$ it is convergent or divergent. The first step to solve this question will be to write the value of ${{a}_{n+1}}$ and ${{a}_{n}}$. On writing the expression in mathematical form we get:

$\Rightarrow {{a}_{n}}=\dfrac{{{x}^{n}}}{{{n}^{2}}+1}$

Now on writing the value of ${{a}_{n+1}}$ , where we will substitute the variable “n” with that of the variable”n+1”. So on doing this we get:

$\Rightarrow {{a}_{n+1}}=\dfrac{{{x}^{n+1}}}{{{\left( n+1 \right)}^{2}}+1}$

The next step will be to find $L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$ , so on substituting the value on the given expression we get:

$\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{\dfrac{{{x}^{n+1}}}{{{(n+1)}^{2}}+1}}{\dfrac{{{x}^{n}}}{{{n}^{2}}+1}} \right|$

For solving the above expression we will cancel out the terms and get:

\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}^{n+1}}\left( {{n}^{2}}+1 \right)}{\left( {{(n+1)}^{2}}+1 \right){{x}^{n}}} \right|\]

\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{x}^{n}}\times x\left( {{n}^{2}}+1 \right)}{\left( {{(n+1)}^{2}}+1 \right){{x}^{n}}} \right|\]

\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{x\left( {{n}^{2}}+1 \right)}{\left( {{(n+1)}^{2}}+1 \right)} \right|\]

To solve it further we will divide the numerator and the denominator with ${{n}^{2}}$ on doing this we get:

\[\Rightarrow L=\displaystyle \lim_{n \to \infty }\left| \dfrac{x\left( \dfrac{{{n}^{2}}}{{{n}^{2}}}+\dfrac{1}{{{n}^{2}}} \right)}{\left( \dfrac{{{(n+1)}^{2}}+1}{{{n}^{2}}} \right)} \right|\]

\[\Rightarrow \displaystyle \lim_{n \to \infty }\left| \dfrac{x\left( \dfrac{{{n}^{2}}}{{{n}^{2}}}+\dfrac{1}{{{n}^{2}}} \right)}{\left( {{\left( \dfrac{n+1}{n} \right)}^{2}}+\dfrac{1}{{{n}^{2}}} \right)} \right|\]

Now on putting the limit in the above expression we get:

\[\Rightarrow L=\left| \dfrac{x\left( 1+\dfrac{1}{\infty } \right)}{\left( 1+\dfrac{1}{\infty } \right)} \right|\]

\[\Rightarrow L=\left| x \right|\]

For convergence the value of \[L<1\]which means the value of $\left| x \right|<1$ to find the value of $x$for which the series becomes convergent or divergent.

According to the ratio test the series will be convergent for $x\le 1$ and will be divergent for $x>1$.

$\therefore $ The series $1+\dfrac{x}{2}+\dfrac{{{x}^{2}}}{5}+\dfrac{{{x}^{3}}}{10}........\dfrac{{{x}^{n}}}{{{n}^{2}}+1}+.......$ will be convergent for $x\le 1$ and will be divergent for $x>1$.

**Note:**To find knowing the convergence and divergence of the series we need solve it using know the concept of ration test which states:

Let the value of $L=\displaystyle \lim_{n \to \infty }\left| \dfrac{{{a}_{n+1}}}{{{a}_{n}}} \right|$ be the condition then:

If $L<1$ , the series is convergent.

If $L>1$ , the series is divergent.

If $L=1$ , then the ratio test does not give any conclusion.

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