# 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,

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

• What if I add the restriction that $w_a+k =< w_{a+1}$ for all a and some fixed k? Would my proposition then hold? I believe that it would, but have trouble formalizing any kind of proof. – heepo Jul 20 at 23:36
• No. Any difference at the beginning isn’t wiped out by adding more and more identical terms to each partial sum. Here’s a simpler example. Let $k=1$ and $w_i=i$. Let $f(n)$ equal $w_n$ except that $w_1=0$. Something approximately like you are thinking about might be true for averages, but not sums. – Steve Kass Jul 21 at 0:24
• Would that mean that the limit as n tends toward infinity of $p_n + p_{n-1} ...$ is not equal to $nln(n) + (n-1)ln(n-1)...$? p_n being the n-th prime, that is. – heepo Jul 21 at 1:27
• or (referring to my last comment) would it be that when $w_n$ tends toward plus or minus infinity that my proposition holds, and $p_n + p_{n-1} + p_{n-2}...$ is an instance of this? – heepo Jul 21 at 1:42
• I suppose it would suffice to say that the sum diverges. – heepo Jul 21 at 7:14

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.