Question

For $p, q, r \in ℝ, r \neq 0$ and $n \in ℕ$ , let
$a_{n} = p^{n} n^{q} {(\frac{n}{n + 2})}^{n^{2}}$ and $b_{n} = \frac{n^{n}}{n! r^{n}} (\sqrt{\frac{n + 2}{n}}) .$
Then, which one of the following statements is TRUE?

$If 1 < p < e^{2} and q > 1, then \sum_{n = 1}^{\infty} a_{n}$ is convergent
$If e^{2} < p < e^{4} and q > 1, then \sum_{n = 1}^{\infty} a_{n}$ is convergent
$If 1 < r < e, then \sum_{n = 1}^{\infty} b_{n}$ is convergent
$If \frac{1}{e} < r < 1, then \sum_{n = 1}^{\infty} b_{n}$ is convergent

Answer: $If 1 < p < e^{2} and q > 1, then \sum_{n = 1}^{\infty} a_{n}$ is convergent

We have $a_{n} = p^{n} n^{q} {(\frac{n}{n + 2})}^{n^{2}} .$ So,

\begin{array}{rcl} \sum_{n = 1}^{\infty} a_{n} & = & \sum_{n = 1}^{\infty} p^{n} n^{q} {(\frac{n}{n + 2})}^{n^{2}} \\ = & \sum_{n = 1}^{\infty} n^{q} {(\frac{n}{n + 2})}^{n^{2}} \cdot p^{n} \end{array}

Here, Series $\sum_{n = 1}^{\infty} a_{n}$ is the power Series in variable $p .$

Now, assume $c_{n} = n^{q} {(\frac{n}{n + 2})}^{n^{2}} .$ Hence,

\begin{array}{rcl} L & = & lim_{n \to \infty} sup {∣ c_{n} ∣}^{\frac{1}{n}} \\ = & lim_{n \to \infty} sup {| n^{q} {(\frac{n}{n + 2})}^{n^{2}} |}^{\frac{1}{n}} \\ = & lim_{n \to \infty} sup | {(n^{q})}^{\frac{1}{n}} {(\frac{n}{n + 2})}^{n} | . \end{array}

We know that $lim_{n \to \infty} {(n^{q})}^{\frac{1}{n}} = 1, \forall q \in ℝ^{+}$ and

\begin{array}{rcl} lim_{n \to \infty} {(\frac{n}{n + 2})}^{n} & = & lim_{n \to \infty} {(\frac{n + 2 - 2}{n + 2})}^{n} \\ = & lim_{n \to \infty} {(1 - \frac{2}{n + 2})}^{n} \\ = & e^{- 2} . \end{array}

Therefore, $L = lim_{n \to \infty} sup ∣ 1 \cdot e^{- 2} ∣ = e^{- 2} .$

Now, the radius of convergence of the Series $\sum_{n = 1}^{\infty} a_{n}$ is

R = \frac{1}{L} = \frac{1}{e^{- 2}} = e^{2} .

Hence, Interval of convergence for $' p^{'}$ of the Series $\sum_{n = 1}^{\infty} a_{n}$ is

∣ p ∣ < R ⟹ ∣ p ∣ < e^{2} ⟹ - e^{2} < p < e^{2} .

Hence, the Series $\sum_{n = 1}^{\infty} a_{n}$ is also converges if $1 < p < e^{2} .$

So, the option (1) is correct and option (2) is wrong.

Now, We have $b_{n} = \frac{n^{n}}{n! r^{n}} (\sqrt{\frac{n + 2}{n}}) .$ So,

\sum_{n = 1}^{\infty} b_{n} = \sum_{n = 1}^{\infty} \frac{n^{n}}{n! r^{n}} (\sqrt{\frac{n + 2}{n}}) .

\begin{array}{rcl} ∴ b_{n + 1} & = & \frac{{(n + 1)}^{n + 1}}{(n + 1)! r^{n + 1}} (\sqrt{\frac{n + 3}{n + 1}}) \\ = & \frac{{(n + 1)}^{n} (n + 1)}{(n + 1) n! r^{n} \cdot r} (\sqrt{\frac{n + 3}{n + 1}}) \\ = & \frac{{(n + 1)}^{n}}{n! r^{n} \cdot r} (\sqrt{\frac{n + 3}{n + 1}}) . \end{array}

Apply the Ratio Test, we get

\begin{array}{rcl} lim_{n \to \infty} \frac{b_{n + 1}}{b_{n}} & = & lim_{n \to \infty} (\frac{\frac{{(n + 1)}^{n}}{n! r^{n} \cdot r} (\sqrt{\frac{n + 3}{n + 1}})}{\frac{n^{n}}{n! r^{n}} (\sqrt{\frac{n + 2}{n}})}) \\ = & lim_{n \to \infty} ({(1 + \frac{1}{n})}^{n} \cdot \frac{1}{r} \cdot \sqrt{\frac{n + 3}{n + 1} \times \frac{n}{n + 2}}) \\ = & \frac{e}{r} . \end{array}

Therefore, the Series $\sum_{n = 1}^{\infty} b_{n}$ converges only if

\frac{e}{r} < 1 ⟹ \frac{r}{e} > 1 ⟹ r > e .

Hence, option (3) and (4) are also wrong.

So, the correct answer is option-1. $\to If 1 < p < e^{2} and q > 1, then \sum_{n = 1}^{\infty} a_{n}$ is convergent