Given the series $$\sum_{n=2}^{\infty} \left( 1-\frac{1}{n^2} \right)^{n^α}$$ I need to study the convergence of the series in correspondence of the variation of $α$.
I've tried to apply all the convergence tests I know, with no luck. I'm kinda new to series exercises. Any help?
If the series converges, the sequence of which you take the sum must converge to 0. So you have
$$\sum\limits_{n=2}^\infty \Big(1-\frac{1}{n^2}\Big)^{n^\alpha} < \infty \Rightarrow \lim\limits_{n\to \infty} \Big(1-\frac{1}{n^2}\Big)^{n^\alpha} = 0 $$
By Bernoulli's inequality we conclude:
$$ \Big( 1-\frac{1}{n^2} \Big)^{n^\alpha} \geq 1+n^\alpha \big(-\frac{1}{n^2} \big) = 1-n^{\alpha-2}$$
Therefore: $$\alpha = 2: \lim\limits_{n \to \infty} \Big( 1-\frac{1}{n^2} \Big)^{n^2} = \lim\limits_{n \to \infty} \Big( 1-\frac{1}{n} \Big)^{n} = \frac{1}{e} > 0$$
$$\alpha < 2: \Big( 1-\frac{1}{n^2} \Big)^{n^\alpha} > 1-n^{\alpha-2} \to 1 \neq 0$$
So the series diverges for $\alpha \leq 2$.
For $\alpha > 2$ we see that, since $\Big( 1-\frac{1}{n} \Big)^n$ is an increasing sequence, $$ \sum\limits_{n=2}^\infty \Big(1-\frac{1}{n^2}\Big)^{n^\alpha} \leq \sum\limits_{n=2}^\infty \Big(\frac{1}{e}\Big)^{n^{\alpha-2}} $$ Now this is looks like a geometric series. If $\alpha - 2 > 1$ we have convergence, since the sequence in the sum is even faster converging towards 0 than in a normal geometric series. If $\alpha -2 =1$ it's just a geometric series. If $1 > \alpha-2 > 0$ we see that this too converges, if we take a look at the answer by ajotatxe in
Does $\sum q^\sqrt n$ converge?
Therefore the series diverges for $\alpha \leq 2$ and converges for $\alpha > 2$.
If $\alpha\leq 2$ the main term does not converge to $0$ as $n\to +\infty$, so the series is divergent. If $\alpha > 2$ $$ \sum_{n\geq 2}\left(1-\frac{1}{n^2}\right)^{n^\alpha} \leq \sum_{n\geq 2}e^{-n^{\alpha-2}} $$ is blatantly convergent by comparison with $\int_{0}^{+\infty}\exp\left(-x^{\alpha-2}\right)\,dx=\Gamma\left(\frac{\alpha-1}{\alpha-2}\right)$.
