Proof of $f(n) + ο(f(n)) = \Theta(f(n))$
Obviously, $f(n)+o(f(n))=\Omega(f(n))$ (clearly, I'm assuming all functions being positive), so you need only to prove that $f(n)+o(f(n))=O(f(n))$. But a function in $o(f(n))$ is definitevely smaller than $f(n)$, so for sufficient large $n$ you have $f(n)+o(f(n))\leq 2f(n)=O(f(n))$.
Actually, in the same way you can easily prove also this stronger claim: $f(n)+O(f(n))=\Theta(f(n))$.

If $h(n)=o(f(n))$, then $\lim_{n\to+\infty} \frac{h(n)}{f(n)}=0$, so there exists $n_0$ such that for all $n\geq n_0$ you have $h(n)\leq f(n)$ (actually, we can prove a stronger statemen, namely that for all $\varepsilon&gt;0$, we have $h(n)&lt; f(n)$, but here we not need it). This means that for all $n\geq n_0$ you also have $f(n)+h(n)\leq 2f(n)$, and then $f(n)+h(n)=O(f(n))$.


If you don't want to explicitely assume positiveness of the functions involved; I suggest to use limits (I like limits).
Remember that $f(n)=O(g(n))$ means
$$
\limsup_{n\to+\infty} \frac{|f(n)|}{g(n)}&lt;+\infty
$$
while
$f(n)=\Omega(g(n))$ means
$$
\liminf_{n\to+\infty} \frac{|f(n)|}{g(n)}&gt;0
$$
(use $|g(n)|$ instead of $g(n)$ if you want to take into account also negative functions). Notice that I'm not assuming that $\lim_{n\to+\infty} \frac{f(n)}{g(n)}$ exists, indeed $\liminf$ and $\limsup$ are used to deal with this case.
Now, as above, let $h(n)$ be an $o(f(n))$, e.g. $\lim_{n\to+\infty} \frac{h(n)}{f(n)}=0$.
Finally
$$
\limsup_{n\to+\infty} \frac{|f(n)+h(n)|}{|f(n)|}&lt;\limsup_{n\to+\infty} \frac{|f(n)|+|h(n)|}{|f(n)|}=\limsup_{n\to+\infty}\left( 1+\frac{|h(n)|}{|f(n)|}\right)=1&lt;\infty
$$
while
$$
\liminf_{n\to+\infty} \frac{|f(n)+h(n)|}{|f(n)|}=\liminf_{n\to+\infty} |f(n)|\cdot\frac{\left|1+\frac{|h(n)|}{|f(n)|}\right|}{|f(n)|}\geq 1&gt;0.
$$
Also observe that in the first limit, we only need $\limsup_{n\to+\infty} \frac{|h(n)|}{|f(n)|}$ to be finite, so $h(n)$ can be also a $O(f(n))$.