Let $(a_n)_{n\geq1}$ a sequence strictly increasing of real positive numbers such that $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}=1$.
Find $$\lim_{n\to\infty} \sum_{k=1}^{n} \frac{a_k}{a_k+a_1+a_2+...+a_n}$$ I know this should be solved using Riemann integration, but my only significant progress was the finding of the partition $$0\leq\frac{a_1}{a_1+...+a_n}\leq\frac{a_1+a_2}{a_1+...+a_n}\leq...\leq\frac{a_1+...+a_n}{a_1+...+a_n}=1$$ for the interval $[0,1]$.
Let $S_n = \sum_{k=1}^{n} {(a_1+a_2+...+a_n)}$; assume we show that $\frac{a_n}{S_n}$ goes to zero as $n$ goes to $\infty$; then if $b_n = \sum_{k=1}^{n} \frac{a_k}{a_k+a_1+a_2+…+a_n}$ is the general term of the sequence above it is obvious that (using $0 < a_k <a_n$): $\frac{S_n}{a_n+S_n} < b_n < 1$ so $b_n$ converges to $1$
Let $m > 1$ an integer and pick $0 < \delta < 1$, s.t $\sum_{k=1}^{m} {\delta^{k}} > \frac{m}{2}$, for example any $\log{\delta} > -\frac{\log 2}{m}$ will do. Since $\lim_{n\rightarrow\infty} \frac{a_{n+1}}{a_n}=1$, we can find a large enough $n_m$ s.t. $\frac{a_n}{a_{n+1}} > \delta$, for $n>n_m$; then if say $n>n_m+m$, $a_{n-q} > a_n\delta^{q}$ for $1 \leq q \leq m$, so $\frac{a_n}{S_n} < \frac{1}{\sum_{k=1}^{m} {\delta^{k}}} <\frac{2}{m}$ and we are done.
