For each $i \in \{1, 2, 3, ...\}$ define $Y_i = X_i^{-\alpha/2}$. Define $Z_k$ by:
$$Z_k = \left[ 1 + \left( \frac{1}{1+\sum_{i=1}^kY_i} \right)^{\alpha/2} \right]^{k+1} $$
Note that $1 \leq Y_i < \infty$ for all $i$. Considering the extreme cases, we get:
$$ \frac{1}{\left[1+\left(\frac{1}{1+k}\right)^{\alpha/2} \right]^{k+1}} \leq \frac{1}{Z_k}\leq 1$$
Hence, for any $\theta \in [1, \infty)$:
\begin{align}
E\left[\frac{1}{Z_k}\right] &\leq \frac{P\left[\sum_{i=1}^kY_i \leq \theta\right]}{\left[1 + \left(\frac{1}{1+\theta}\right)^{\alpha/2}\right]^{k+1}} + P\left[\sum_{i=1}^kY_i > \theta\right]\\
E\left[\frac{1}{Z_k}\right] &\geq
\frac{P\left[\sum_{i=1}^kY_i\leq \theta\right]}{\left[1+\left(\frac{1}{1+k}\right)^{\alpha/2} \right]^{k+1}} + \frac{P\left[\sum_{i=1}^kY_i > \theta\right]}{\left[1+\left(\frac{1}{1+\theta}\right)^{\alpha/2} \right]^{k+1}}
\end{align}

### Claim:

With probability 1 we have:
$$ \lim_{k\rightarrow\infty} \frac{1}{Z_k} = \left\{ \begin{array}{ll}
0 &\mbox{ if $0<\alpha < 2$} \\
1 & \mbox{ if $\alpha \geq 2$}
\end{array}\right.
$$
Since $0 \leq \frac{1}{Z_k}\leq 1$ for all $k$, the above claim together with the bounded convergence theorem immediately implies:
$$ \lim_{k\rightarrow\infty} E\left[\frac{1}{Z_k}\right] = \left\{ \begin{array}{ll}
0 &\mbox{ if $0<\alpha < 2$} \\
1 & \mbox{ if $\alpha \geq 2$}
\end{array}\right.
$$

### Proof of Claim:

We have:
\begin{align}
\log(Z_k) &= (k+1)\log\left(1 + \left( \frac{1}{1+\sum_{i=1}^kY_i} \right)^{\alpha/2} \right) \\
&= (k+1)\log\left( 1+ \left(\frac{1/k}{1/k+A_k}\right)^{\alpha/2} \right)
\end{align}
where $A_k$ is defined $A_k = \frac{1}{k}\sum_{i=1}^k Y_i$ and $A_k\rightarrow E[Y_1]$ with prob 1 by the law of large numbers. Using the inequality:
$$ \frac{x}{1+x} \leq \log(1+x) \leq x \quad \forall x > -1 $$
we get:
$$(k+1)\frac{\left(\frac{1/k}{1/k+ A_k}\right)^{\alpha/2}}{1+\left(\frac{1/k}{1/k+ A_k}\right)^{\alpha/2}} \leq \log(Z_k) \leq (k+1)\left(\frac{1/k}{1/k+ A_k}\right)^{\alpha/2}$$

If $0 < \alpha/2 < 1$ then $0<E[Y_1]<\infty$ and $A_k\rightarrow E[Y_1]$ with prob 1. We get $\log(Z_k) \rightarrow \infty$ with prob 1. Hence, $Z_k\rightarrow\infty$ and $1/Z_k\rightarrow 0$ with prob 1.

If $\alpha/2 \geq 1$ then $E[Y_1]=\infty$ and $A_k\rightarrow \infty$ with prob 1. So $\log(Z_k)\rightarrow 0$ and so $Z_k\rightarrow 1$ with prob 1. $\Box$