Let $X_1,\ldots,X_n$ be independent and identically distributed random variables and assume that the expectation of $X_1$ is finite. For $k \le n$ find:
$$\mathbb{E}\frac{\sum\limits_{i=1}^k X_i}{\sum\limits_{i=1}^n X_i}$$
so I think I can separate it like this:
$$\frac {\mathbb{E}\sum\limits_{i=1}^k X_i}{\mathbb{E} \sum\limits_{i=1}^n X_i}$$ Is that right?
Then if $k = n$
$$\frac {\mathbb{E}\sum\limits_{i=1}^n X_i}{\mathbb{E} \sum\limits_{i=1}^n X_i} =1 $$
if $k>n$, I think I can write this:
$$\mathbb{E}\sum\limits_{i=n}^k X_i$$
I am really not sure if this is all right, or if I even finished solving this. Any help would be appreciated.
$\newcommand{\E}{\operatorname{E}}$ $$ \E\frac{\sum\limits_{i=1}^k X_i}{\sum\limits_{i=1}^n X_i} = \E\left(\frac{X_1}{\sum\limits_{i=1}^n X_i}\right) + \cdots + \E\left(\frac{X_n}{\sum\limits_{i=1}^n X_i}\right). \tag 1 $$ Since they're i.i.d., every term on the right has the same value, so it's $$ k\E\left(\frac{X_1}{\sum\limits_{i=1}^n X_i}\right). \tag 2 $$ If $k=n$, then the expression on the left in $(1)$ is just $\E 1 =1$, but it must be the expression in $(2)$ with $k=n$. Therefore the value of the expected value in $(2)$ must be $1/n$, and the value of the expression $(2)$ as a whole must therefore by $k/n$.
However, it is not generally true that $\E\dfrac U V= \dfrac{\E U}{\E V}$. Independence does let you conclude that $\E\dfrac U V = \left( \E U\right)\left( \E \dfrac 1 V\right)$ if $U$, $V$ are independent, but notice that $X_1$ and $\sum\limits_{i=1}^n X_i$ are not independent, so that cannot be used here. And $\E\dfrac 1 V$ is not the same as $\dfrac 1 {\E V}$.