Let $X = (X_1, ... X_n)$ consist of $n \ iid$ Binomial rvs $X_i, s.t. X_i\sim \operatorname{Bin}(m, \theta), m \in \mathbb{N}, $ unknown $ \theta \in (0,1)$
Show that if $\hat\theta(X) := \frac{1}{nm+m}\sum_{i=1}^nX_i$ , then $(\hat\theta_n(X))_{n\in \mathbb{N}}$ is a consistent sequence of estimators.
I know that if we find the bias and variance, then if
$\lim_{n \rightarrow \infty} \operatorname{bias}(\hat\theta_n(X), \theta) \rightarrow 0$ and $\lim_{n \rightarrow \infty} \operatorname{Var}_\theta(\hat\theta_n(X)) \rightarrow 0$, then we have a consistent sequence of estimators.
Hence, I have attempted to calcuate these:
(Updated answer, using variance answer from below, advice for bias):
Variance: $\operatorname{Var}_\theta(\frac{1}{nm+m}\sum_{i=1}^nX_i) = \frac{1}{(nm+m)^2}\operatorname{Var}_\theta(\sum_{i=1}^nX_i)$
$=\frac{1}{n^2m^2 + 2nm^2 + m^2}\sum_{i=1}^n \operatorname{Var}_\theta(X_i)$
$=\frac{1}{n^2m^2 + 2nm^2 + m^2} n \operatorname{Var}(X_1)$ - since we have iid rvs
$=\frac{n}{n^2m^2 + 2nm^2 + m^2} m\theta(1-\theta)$
$=\frac{nm\theta(1-\theta)}{n^2m^2 + 2nm^2 + m^2}$
$=\frac{n\theta(1-\theta)}{n^2m + 2nm + m}$
So, $lim_{n \rightarrow \infty}(\frac{n\theta(1-\theta)}{n^2m + 2nm + m})$ = $lim_{n \rightarrow \infty}(\frac{\frac{\theta(1-\theta)}{n}}{m+\frac{2m}{n}+\frac{m}{n^2}}) = 0$
Then Bias: \begin{align} \mathbb{E}_\theta(\frac{1}{nm+m}\sum_{i=1}^n X_i) - \theta &=\frac{1}{nm+m}\mathbb{E}_\theta(\sum_{i=1}^n X_i)-\theta\\ &=\frac{1}{nm+m}n\mathbb{E}(X_1) - \theta \mbox{ since we have iid rvs}\\ &=\frac{nm\theta}{nm+m}-\theta\\ &=\theta(\frac{nm}{nm+m}-1)\\ &=\theta(\frac{mn}{m(n+1)}-1)\\ &=\theta(\frac{n}{n+1}-1)\\ &=\frac{n\theta}{n+1} - \theta\\ &=\frac{n\theta-\theta}{n+1} \end{align} Then, $\lim_{n \rightarrow \infty}(\frac{n\theta-\theta}{n+1}) = \lim_{n \rightarrow \infty} \frac{\frac{\theta}{n}-\frac{\theta}{n^2}}{\frac{1}{n}+\frac{1}{n^2}}=0$
And hence we have a consistent sequence of estimators
For the variance, I divide $n^2$ in both the numerator and the denominator. \begin{align} \lim_{n \to \infty} \frac{n \theta(1-\theta)}{n^2m+2nm+m} = \lim_{n \to \infty}\frac{\frac{\theta(1-\theta)}{n}}{m+\frac{2m}{n}+\frac{m}{n^2}}=0 \end{align}
For the bias, many of your equal signs should be minus sign instead. The last few lines of your working doesn't make sense. But you should use a similar trick by dividing $n$ in both the numerator and denominator to get a conclusion.
Edit:
$$\lim_{n \to \infty} \frac{n \theta}{n+1}-\theta=\lim_{n \to \infty}\frac{\theta}{1+\frac1n}-\theta=\theta - \theta = 0$$