I decided to try this with a simple sum I made
$$\sum_{i=1}^{n}{\frac{x}{i}} = 3$$
I know the basic properties of sums, and I tried solving this with the properties; $\sum_{i=1}^{n}{i} = \frac{1}{2}n(n+1)$ and $\sum_{i=1}^{n}{c} = nc$
Doing so gave me this:
$$\frac{nx}{\sum_{i = 4}^{n}{i}} =3$$ Which I simplified into $$\frac{nx}{\frac{1}{2}n(n+1)} = 3$$ and then finally, $$x = \frac{3}{2n}n(n+1)$$ and when refined; $$x = \frac{3n + 3}{2}$$
Which I tried to solve using $n=4$, but got $\frac{30}{8}$ which when plugged into the same sum with $n=4$ returned $\frac{25^2}{4^2}$ for reasons I do not know. I do know that $\frac{125}{16} \ne 3$, so where did I go wrong, and how do I solve this correctly? I'm not a calculus student, I'm actually an algebra II student, so please keep it to simple stuff like limits and sums, which is about all I understand.
It is not true that $$\sum_{i=1}^n \dfrac{x}{i} = \dfrac{nx}{\sum_{i=1}^n i}$$
However, you can just factor out the $x$ since it does not depend on $i$ and find $$x= \dfrac{3}{\sum_{i=1}^n \frac{1}{i}}$$
The numbers $\sum_{i=1}^n \frac{1}{i}$ are known as harmonic numbers. I am unaware of any 'nice' formula for them in general - at least any that make the formula already nicer than it is.