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 $j$?

Question

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?

Answer 1

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,

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

Answer 2

I assume you mean $\lim_{n\to\infty} w_n-f(n)=0$, because $\lim_{n\to\infty} w_n=f(n)$ doesn't make sense.

Nothing more false. Mainly, a possibly large difference between $w_1$ and $f(1)$ translates tout court in an equally large difference between the partial sums, which (in principle and sometimes in point of fact) is not improved by the tailing terms being close by.