Almost sure convergence of $ \frac{ X_n +1}{n } $ Given that $P( \ X_n = n^2 -1 ) = n^{-2} , n \ge 1  $
$P( \ X_n = -1 ) = 1- \frac{1}{n^2}    $
$ Show $
$ \sum_{n =1}^\infty \frac{ X_n +1}{n }    $  converges almost sure and
$ \  S_n = X_1+...X_n , $
$show  $
$ \frac{\ S_n }{n} \rightarrow -1 $
My attempt
For $P( \ X_n = n^2 -1 ) = n^{-2} $
$ \sum_{n =1}^\infty \frac{ X_n +1}{n } = \sum_{n =1}^\infty \frac{ 1+ n^2}{n^3 }  $
$ \sum_{n =1}^\infty [ \frac{ 1}{n^3 } + \frac{1}{n} ]  $
Which $ \rightarrow 1 $  as $ n \rightarrow   \infty $
For $P( \ X_n = -1 ) = 1- \frac{1}{n^2} $
$ \sum_{n =1}^\infty \frac{ X_n +1}{n } = \sum_{n =1}^\infty \frac{ -1 }{n^3 }  $
Which $ \rightarrow -1 $  as $ n \rightarrow   \infty $
Can anyone please tell if I am going the right way ??
Please let me know how to proceed  for $ \ S_n $ part , I tried but I am not getting it -1

First, you should specify that the random variables $X_n$ are independent, otherwise this problem isn't well-defined.
To make the problem easier, note that we can rewrite $Y_n = X_n+1$ so that the problem becomes

Given random independent variables $Y_1, Y_2, \ldots, $ where $Y_n\sim \operatorname{Bernoulli}(1/n^2)\cdot n^2$, show that $\sum_{i=1}^n \frac{Y_i}{i}$ converges almost surely as $n\to\infty$ and that $\frac{1}{n}\sum_{i=1}^n Y_i\to 0$.

The first half is true by Borel-Cantelli since that tells us that almost surely, only a finite number of the terms are nonzero. The second one follows by the same reasoning. Almost surely, only a finite number of terms are nonzero, so since $\sum_{i=1}^\infty Y_i$ is almost-surely finite, we must have $\frac{1}{n}\sum_{i=1}^n Y_i\to 0$.
