Convergence in Probability In the following question, I was able to compute part (i), and part (ii), but I am finding hard time to compute with part (iii).

So far I have computed the MLE of $\theta$, which is $\frac{\sum_{i=1}^n Y_i -n}{\sum_{i=1}^n Y_i - M} $
I was trying to use the following concept for convergence

but I am finding it hard to proceed. Any help would be highly appreciated.
 A: Convergence in probability means
$$\operatorname{plim}(\hat \theta) = \theta$$
The $\operatorname{plim}(\hat \theta)$ is straightforward to compute:
$$\operatorname{plim}(\hat \theta) = \operatorname{plim}\left(\frac{\sum_{i=1}^n Y_i -n}{\sum_{i=1}^n Y_i - M}\right)$$
and multiplying numerator and denominator by $\frac 1n$ we obtain, using the continuous transformation mapping and the LLN
$$\operatorname{plim}(\hat \theta) = \frac {E(Y) - 1}{E(Y)-P(Y=r+1)}$$
Note that 
$$P(Y=r+1) = \operatorname{plim}(M/n) = P(X>r) = 1-P(X\le r)$$  
and so 
$$\operatorname{plim}(\hat \theta) = \frac {E(Y) - 1}{E(Y)-1+P(X\le r)}$$
which provides a nice intuition. Now since you have computed the pmf of $Y$ you should be able to show that 
$$\operatorname{plim}(\hat \theta) = \frac {E(Y) - 1}{E(Y)-1+P(X\le r)} = \theta$$
ADDENDUM
Note that what is denoted $p$ in the textbook formulation of the geometric distribution (the probability of "success"), in your formulation it is $1-\theta$. So we have
$$P(X\le r) = 1-\theta^{\,r}$$
So we want to check whether
$$\operatorname{plim}(\hat \theta) = \frac {E(Y) - 1}{E(Y)-\theta^{\,r}} =?\; \theta \Rightarrow (1-\theta)E(Y) =?\; 1-\theta^{\,r+1} \qquad [A]$$
The variable $Y$ can be written, using indicator functions, as
$$Y = X\cdot I_{\{X\le r\}} + (r+1)(1-I_{\{X\le r\}})$$
So, using the properties of indicator functions
$$E(Y) =E\left( X\cdot I_{\{X\le r\}}\right) + (r+1)\left(1-P(X\le r)\right) = E\left( X\cdot I_{\{X\le r\}}\right)+(r+1)\theta^{\,r}$$
Consider now the variable $Z=X\cdot I_{\{X\le r\}}$.This variable takes the values of $X$ when $X\le r$ and the value $0$ otherwise. So we have immediatelly
$$E(Z) = E\left( X\cdot I_{\{X\le r\}}\right) = \sum_{k=1}^rk(1-\theta)\theta^{\,k-1}= \frac {1-\theta}{\theta}\sum_{k=1}^rk\theta^{\,k}$$
The value of this finite sum is (look it up, for example here)
$$\sum_{k=1}^rk\theta^{\,k} =  \theta\frac{1-(r+1)\theta^{\,r}+r\theta^{\,r+1}}{(1-\theta)^2}$$
Simplifying we obtain
$$E(Z) = E\left( X\cdot I_{\{X\le r\}}\right) =\frac{1-(r+1)\theta^{\,r}+r\theta^{\,r+1}}{1-\theta} $$
So the LHS of eq. $[A]$ becomes
$$(1-\theta)E(Y) = 1-(r+1)\theta^{\,r}+r\theta^{\,r+1}+(1-\theta)(r+1)\theta^{\,r}$$
$$=1-(r+1)\theta^{\,r}+r\theta^{\,r+1}+(r+1)\theta^{\,r}-(r+1)\theta^{\,r+1}$$
Simplifying we obtain 
$$(1-\theta)E(Y)=1-\theta^{\,r+1}$$
which is exactly what we wanted to verify, in order to show that the estimator is consistent.
On the side, note the interesting thing here that the above implies that $E(Y) = 1+\theta^2+...+ \theta^{\,r}$, which should be intuitive, since as $r \rightarrow \infty$ (no failure), $E(Y) \rightarrow \frac {1}{1-\theta}=E(X)$.
