If I understand $$\sum _{n\leq x}a_n(1-n/x)\hspace {5mm}\text {or}\hspace {5mm}\sum _{n\leq x}a_n\log (x/n)$$ then I understand $$\sum _{n\leq x}a_n$$ by writing $$\sum _{n\leq x}a_n=\frac {x+h}{h}\sum _{n\leq x+h}a_n(1-n/(x+h))-\frac {x}{h}\sum _{n\leq x}a_n(1-n/x)+\mathcal O(h)$$ and $$\sum _{n\leq x}a_n\int _x^{x+h}\frac {dt}{t}=\int _x^{x+h}\frac {1}{t}\left (\sum _{n\leq t}a_n\right )dt+\mathcal O(h)=\sum _{n\leq x+h}a_n\log ((x+h)/x)=\sum _{n\leq x+h}a_n\log ((x+h)/n)-\sum _{n\leq x}a_n\log (x/n)+\mathcal O(h).$$ Another common choice of weight is $e^{-n/x}$. But how do I link $$\sum _{n=1}^\infty a_ne^{-n/x}$$ sum to the unweighted sum?
(If I just want an upper bound I'm ok, but let's say I want an asymptotic. Also notice partial summation doesn't work, because the end-point change (the weights are no longer smooth)).
I've tried a few things like seeing what Taylor expansions of things like $e^{n/(x+h)$ give but I think really I don't know what I'm doing.
It depends on how much you assume for $a_n$. If $a_n\ge0$, then there is
$$ \sum_{n=1}^\infty a_ne^{-n/x}\ge\sum_{n\le x}a_ne^{-n/x}\ge e^{-1}\sum_{n\le x}a_n. $$