# Find $\lim\limits_{n\rightarrow\infty} \sum\limits_{k=1}^{n} \frac{a_k}{a_k+a_1+a_2+...+a_n}$

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 $$0 < a_k ): $$\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.