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I've been having trouble checking whether this sequence converges or not:

$$\sum_{n=0}^{\infty} \frac{n-3}{n+2}^\left({n^2-n}\right)$$

At a first glance I thought I should try the root test but that didn't work (maybe I did something wrong?)

Some help?

Thanks.

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I think the root test works $\mathop {\lim }\limits_{n \to \infty } \sqrt[n]{{{{\left( {\frac{{n - 3}}{{n + 2}}} \right)}^{{n^2} - n}}}} = \mathop {\lim }\limits_{n \to \infty } {\left( {\frac{{n - 3}}{{n + 2}}} \right)^{n - 1}} = {e^{ - 5}}$

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Since: $$\left(1-\frac{1}{n}\right)^n \leq \frac{1}{e}, \tag{1}$$ we have: $$\sum_{n\geq 1}\left(1-\frac{1}{n}\right)^{n^2}\leq \sum_{n\geq 1}\frac{1}{e^n} = \frac{1}{e-1}.\tag{2}$$ With minor adjustments (I leave it to you to find them) this proves that your series is converging.

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