Question

Find the positive values of $p$ for which $\sum {n{{\left( {1 + {n^2}} \right)}^p}}$ from $\left[ {2,\infty } \right)$ converges?

Answer 1

This problem deals with finding the positive values of $p$, for which the given expression converges within the given interval. So check whether the given expression converges or not there are few tests in mathematics, which concludes according to the tests. But here we are going to apply the limit to the expression to check the convergence on the given interval.

Complete step by step solution:
Let the expression $n{\left( {1 + {n^2}} \right)^p}$ be ${a_n}$, as shown below:
$\Rightarrow {a_n} = n{\left( {1 + {n^2}} \right)^p}$
$\therefore \sum {n{{\left( {1 + {n^2}} \right)}^p}} = \sum {{a_n}}$
The necessary condition for $\sum {{a_n}}$ to converge is that first the term ${a_n} > 0$, it should be greater than zero.
Also the limit of ${a_n}$ should be equal to zero, as shown below:
$\Rightarrow \mathop {\lim }\limits_n {a_n} = 0$
But we know that the value of the expression $\left( {1 + {n^2}} \right)$ is always greater than 1, which is given below:
$\Rightarrow \left( {1 + {n^2}} \right) > 1$, for any value of $p > 0$
$\Rightarrow \mathop {\lim }\limits_n n{\left( {1 + {n^2}} \right)^p} = \infty$
Here this series is not convergent but rather divergent.

Using the test of convergence to find the values of $p$, for which the series is convergent, as shown:
Let $f\left( x \right) = {a_n}$, finding the integral of $f\left( x \right)$ as shown below:
$\Rightarrow f\left( x \right) = x{\left( {1 + {x^2}} \right)^p}$
$\Rightarrow \int {f\left( x \right)} dx = \int\limits_1^\infty {x{{\left( {1 + {x^2}} \right)}^p}dx}$ , this integral should converge.
Substituting the value of ${x^2} = t$, so $2xdx = dt$
$\Rightarrow \int\limits_1^\infty {x{{\left( {1 + {x^2}} \right)}^p}dx} = \dfrac{1}{2}\int\limits_1^\infty {{{\left( {1 + t} \right)}^p}} dt$
$\Rightarrow \dfrac{1}{2}\left. {\dfrac{{{{\left( {1 + {x^2}} \right)}^{p + 1}}}}{{p + 1}}} \right|_1^\infty = \dfrac{1}{2}\left. {\dfrac{{\left( {1 + {x^2}} \right){{\left( {1 + {x^2}} \right)}^p}}}{{p + 1}}} \right|_1^\infty$
Here for $x = 1$, the integral becomes, as shown below:
$\Rightarrow F\left( 1 \right) = \dfrac{{{2^p}}}{{p + 1}}$

Consider the limit as shown below:
$\Rightarrow \mathop {\lim }\limits_{x \to + \infty } {\left( {1 + {x^2}} \right)^{p + 1}}$
So this limit converges only for $\left( {p + 1} \right) < 0$ or $p < \- 1$.
So the series is convergent only for $p < \- 1$.

Note: If given an expression is $\dfrac{{a\left[ n \right]}}{{b\left[ n \right]}}$, and the limit of $\dfrac{{a\left[ n \right]}}{{b\left[ n \right]}}$ is positive, then the sum of $a\left[ n \right]$ converges if and only if the sum of $b\left[ n \right]$ converges.
If the limit of $\dfrac{{a\left[ n \right]}}{{b\left[ n \right]}}$ is zero, and the sum of $b\left[ n \right]$ converges, the sum of $a\left[ n \right]$ also converges.
If the limit of $\dfrac{{a\left[ n \right]}}{{b\left[ n \right]}}$ is infinite, and the sum of $b\left[ n \right]$ diverges, then the sum of $a\left[ n \right]$ also diverges.