I am asked to find the value of:
$$\lim_{n \to \infty} \sum^n_{i=1}f\left(\frac{(\sum^i_{k=0}\alpha_k)^2}{(\sum^n_{k=0}\alpha_k)^2}\right)\frac{\alpha_i}{\sum^n_{k=0} \alpha_k}$$
Where $(\alpha_i)$ is a sequence of real numbers such that $\forall i \in \mathbb{N}\;\;\; 0 < \alpha_i < 1$, $\sum^\infty_{i=0}\alpha_i$ is a divergent sum, and $f$ is a continuous function on $\mathbb{R}$.
Usually, the way I would do this is to fiddle around with it until I get it into this form:
$$\lim_{n \to \infty} \frac{b-a}{n} \sum_{i=1}^n f\left (a + \frac{i}{n} (b-a) \right)$$
Then the answer is just $\int^b_a f(x) dx$. But I don't see any way to do this. Is there some algebraic trick I'm missing to get this into the right form?
With fixed $n$, let $x_0=0$ and $x_i=\dfrac{\sum_{k=1}^i\alpha_k}{\sum_{k=1}^n\alpha_k}$, $1\le i\le n$. Then $\{x_0,\dots,x_n\}$ is a partition of $[0,1]$.
Moreover $$ x_i-x_{i-1}=\frac{\alpha_i}{\sum_{k=1}^n\alpha_k}<\frac{1}{\sum_{k=1}^n\alpha_k},\quad 1\le i\le n. $$ Since $\sum\alpha_k$ is divergent, $$ \sup_{i\le i\le n}(x_i-x_{i-1})\to0\text{ as }n\to\infty. $$ The sum can be rewritten as $$ \sum_{i=1}^nf(x_i^2)(x_i-x_{i-1}), $$ which is a Riemann sum for $\int_0^1f(x^2)\,dx$.