<s>[INST] Show $\frac{n+1}{2n^2+5} \in \Theta (\frac{1}{n})$ using definition Show $\frac{n+1}{2n^2+5} \in \Theta (\frac{1}{n})$ using definition
First showing the upper bound 
$\frac{n+1}{2n^2+5} \in O (\frac{1}{n})$
$\frac{n+1}{2n^2+5} \leq \frac{2n}{2n^2} = \frac{1}{n}$
for $n \geq 1$
now showing the lowerbound
$\frac{n+1}{2n^2+5} \in \Omega (\frac{1}{n})$
$\frac{n+1}{2n^2+5} > \frac{n}{2n^2+5} \geq \frac{n}{2n^2 + 5n^2} = \frac{n}{7n^2} = \frac{1}{7n}$
for $n \geq 1$
hence, $\frac{n+1}{2n^2+5} \in \Theta (\frac{1}{n})$
is this right?
 [/INST] Yes, you have shown found constant $c_1$ and $c_2$ such that 
$$ \frac{c_1}{n}\le\frac{n+1}{2n^2+5} \le \frac{c_2}{n}.$$
You have proven what you want.
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