A fair $n(n > 1)$ faces die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$, then $n$ is equal to ____________.

In this case, the success is defined as getting a number less than $n$ when rolling an $n$-sided die. The probability of success, $p$, in each roll is then $(n-1)/n$, and the probability of failure, $q = 1 - p$, is $1/n$. <br/><br/>$$ \text { Mean }=\sum\limits_{i=1}^{\infty} p_i x_i=1 \cdot \frac{n-1}{n}+\frac{2}{n} \cdot\left(\frac{n-1}{n}\right)+\frac{3}{n^2}\left(\frac{n-1}{n}\right)+\ldots $$ <br/><br/>$\frac{n}{9}=\left(1-\frac{1}{n}\right) S$ ......(1) <br/><br/>where <br/><br/>$$ \begin{aligned} & S=1+\frac{2}{n}+\frac{3}{n^2}+\frac{4}{n^3}+\ldots \\\\ & \frac{1}{n} S=\frac{1}{n}+\frac{2}{n^2}+\frac{3}{n^3}+\ldots \\\\ & ----------------\\\\ & \left(1-\frac{1}{n}\right) S=1+\frac{1}{n}+\frac{1}{n^2}+\frac{1}{n^3}+\ldots \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \Rightarrow \left(1-\frac{1}{n}\right) S=\frac{1}{1-\frac{1}{n}} \\\\ & \Rightarrow \frac{n}{9}=\left(1-\frac{1}{n}\right) \times \frac{1}{\left(1-\frac{1}{n}\right)^2}=\frac{n}{n-1} \end{aligned} $$ <br/><br/>$\Rightarrow$ n = 10