Question

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

Answer 1

We have given an infinite series and, we have to find the positive values of $p$ for which the given series converges.
An infinite series is said to be convergent if its limit exists that means the sum of the term is finite that is $\sum\limits_{n \to \infty } {{a_n} = S}$ where $S$ is a finite value.
An infinite series is said to be divergent if its limit does not exist, that means the sum of the term is infinite that is $\sum\limits_{n \to \infty } {{a_n} = \infty }$.

Complete step by step solution:
Given summation is $\sum {n{{\left( {1 + {n^2}} \right)}^p}}$so the ${n^{th}}$ term of the series is
${a_n} = n{\left( {1 + {n^2}} \right)^p}$ .
Now we use the integral test of convergence to find the values of $p$ for which the series converges. Let us consider the function $f\left( x \right)$ such that $f\left( n \right) = {a_n}$ so $f\left( x \right) = x{\left( {1 + {x^2}} \right)^p}$ .
Now the given series converges if the value of $\int\limits_2^\infty {f\left( x \right)dx}$ is finite, so first we simplify the above integral.

Integral is given as $\int\limits_2^\infty {f\left( x \right)dx} = \int\limits_2^\infty {x{{\left( {1 + {x^2}} \right)}^p}dx}$.
To simplify the above integral, we use integration by substitution method. In above integral we substitute
$1 + {x^2} = t \\\ \Rightarrow 2xdx = dt \\\$
For changing the limits from $x$ to $t$ , when $x = 2 \Rightarrow t = 5$ and when $x = \infty \Rightarrow t = \infty$
Substituting the values in above integral, we get
$\Rightarrow \int\limits_5^\infty {{{\left( t \right)}^p}\dfrac{{dt}}{2}}$

Integrate the above integral using power rule, we get
$\Rightarrow \left[ {\dfrac{{{{\left( t \right)}^{p + 1}}}}{{2\left( {p + 1} \right)}}} \right]_5^\infty$
Step 4: Now the lower limit is finite for the finite values of $p$ except for $- 1$ so $p \ne - 1$ but the upper limit is always having infinite value for positive values of $p$ so the given series does not converge for the positive values of $p$ .

Note: Power rule of integration is $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} + C$.
In integration by substitution method of integration, substitute $t = f\left( x \right)$ such that its derivative is also present in the given integral.