if $f(x) \sim g(x)$ is $ \sum f(k) \sim \sum g(k)$ if $f(x) \sim g(x)$ as $x \to \infty$ then is $\sum_{k=1}^N f(k) \sim \sum_{k=1}^N g(k)$ as $N \to \infty$? Intuitively, i should think so because as $k$ gets larger $f$ and $g$ get closer so it should 'even out'?
 A: $f(x)\sim g(x)$ as $x\to\infty$ means
$$
\lim_{x\to\infty}\frac{f(x)}{g(x)}=1
$$
However, suppose that $f(x)=\frac1{x(x+1)}$ and $g(x)=\frac1{x(x+2)}$, then $f(x)\sim g(x)$, but
$$
\sum_{k=1}^Nf(k)=1-\frac1{N+1}
$$
while
$$
\sum_{k=1}^Ng(k)=\frac34-\frac{2N+3}{2(N+1)(N+2)}
$$
So
$$
\sum_{k=1}^Nf(k)\not\sim\sum_{k=1}^Ng(k)
$$

However, if the the sum of the functions tend to monotonically to $\infty$, then choose an $\epsilon\gt0$ and then find an $N_\epsilon$ so that for $x\ge N_\epsilon$,
$$
\left|\frac{f(x)}{g(x)}-1\right|\le\epsilon
$$
Then for $k\ge N_\epsilon$,
$$
|f(k)-g(k)|\le\epsilon|g(k)|\implies\left|\sum_{k=N_\epsilon}^Nf(k)-g(k)\right|\le\epsilon\left|\sum_{k=N_\epsilon}^Ng(k)\right|
$$
Thus, for $N\ge N_\epsilon$,
$$
\left|\frac{\displaystyle\sum_{k=N_\epsilon}^Nf(k)}{\displaystyle\sum_{k=N_\epsilon}^Ng(k)}-1\right|\le\epsilon
$$
Both $\displaystyle\sum_{k=N_\epsilon}^Nf(k)$ and $\displaystyle\sum_{k=N_\epsilon}^Ng(k)$ are unbounded, yet both $\displaystyle\sum_{k=1}^{N_\epsilon-1}f(k)$ and $\displaystyle\sum_{k=1}^{N_\epsilon-1}g(k)$ are fixed, therefore,
$$
\lim_{N\to\infty}\left|\frac{\displaystyle\sum_{k=1}^Nf(k)}{\displaystyle\sum_{k=1}^Ng(k)}-1\right|=0
$$
That is, if the sums monotonically diverge,
$$
\sum_{k=1}^Nf(k)\sim\sum_{k=1}^Ng(k)
$$
A: This is true when : both functions are not $0$, one of the functions is positive (say $f$) and $\sum f(n) = \infty$
