Does $w_n = f(n)$ for sufficiently large $n$ $\implies \sum_{i=1}^{j}w_i = \sum_{i=1}^{j}f(i)$ for sufficiently large $j$? Let $w_n$ be a real-valued (EDIT: strictly increasing or decreasing) infinite sequence and let $w_n \approx f(n)$ for sufficiently large $n$. Then: $$\sum_{i=1}^{j}w_i = \sum_{i=1}^{j}f(i)$$ For sufficiently large $j$. Is this true? My intuition would tell me yes, as the terms past the $n$-th term for sufficiently large $n$, of which there are infinitely many, would make any terms before that $n$-th term which are not approximated negligible; but certainly, there can exist arbitrarily many terms $w_s$ such that $w_s \not\approx f(s)$, so if the above equality is true, how can it be proven?

Providing a specific example in line with Gae.’s answer, let $f(n)={1\over 10^n}$, and let $w$ be the sequence $47, {1\over100}, {1\over1000}, {1\over10000},\dots$. Then the hypotheses of your question are satisfied. However,

$$\sum_{i=1}^{j}w_i = 47.0111\dots{\mbox{ ($j-1$ i’s), but}}$$ $$\sum_{i=1}^{j}f(i)=0.111\dots \mbox{ ($j$ i’s).}$$