If we have an i.i.d random variable $X_i$ with mean variance $(\mu, \sigma^2)$.
By Law of Large number, we have $\bar{X}\rightarrow^p \mu$.
But can we use Law of large number as well and claim that the expression:
$$\frac{\sum_{i=1}^n X_i}{n+1} \rightarrow^p \mu$$ as well?
I tried the following manipulation:
$\frac{\sum_{i=1}^n X_i}{n+1}= \frac{n}{n+1} \frac{\sum_{i=1}^n X_i}{n}$ and claim by Law of Large number,$\frac{\sum_{i=1}^n X_i}{n} \rightarrow^p \mu$ and $\frac{n}{n+1} \rightarrow 1$ and thus it converges to $\mu$.
Is this approach correct? I couldn't find any theorem supporting such manipulation though.
Yes, your approach is correct. This manipulation is justified by Slutsky's theorem. See the article If $a_n \rightarrow 1$, does $X_n \overset{p}{\rightarrow} X$ implies $a_n X_n \overset{p}{\rightarrow} X$? for a reference to, and discussion about, Slutsky's theorem (and also a proof).