<s>[INST] Examining whether $\sum \limits_{n=0}^\infty\frac{(-1)^{n+1}}{5n+1}$ is convergent, absolute convergent or divergent Everything in red is edited
To show, that the series is convergent we show at first, that $\color{red}{\lim \limits_{n \to \infty} \left(\dfrac{1}{5n+1}\right)}=0$.
$\color{red}{\lim \limits_{n \to \infty} \left(\dfrac{1}{\underbrace{5n+1}_{1/\infty}}\right)=\lim \limits_{n \to \infty} \left(\dfrac{1}{n}\right)}=0 \implies a_n>0$
Leibniz criterion $\sum \limits_{n=0}^\infty (-1)^{n+1}\cdot \underbrace{\dfrac{1}{5n+1}}_{a_n}$
We still need to show, that $a_n$ is monotonic decreasing: 
\begin{align}
a_{n}&\ge a_{n+1}\\
\color{red}{\frac1{5n+1}}&\color{red}{\ge\frac1{5n+6}\iff 5n+6\ge5n+1\iff 6\ge 1\;\checkmark}
\end{align} $\implies$ monotonic decreasing.
$\implies$ The series is convergent.
To prove, that the series is absolute convergent, we show that $\sum \limits_{n=0}^\infty \left|\dfrac{(-1)^{n+1}}{5n+1}\right|$ is converging.
\begin{align}
\sum \limits_{n=0}^\infty \left|\dfrac{(-1)^{n+1}}{5n+1}\right|&=\sum \limits_{n=0}^\infty \dfrac{\mid (-1)^{n+1}\mid }{\mid 5n+1\mid}\\
&=\sum \limits_{n=0}^\infty \frac{1}{5n+1}\\
&\ge\color{red}{\sum \limits_{n=0}^\infty \frac{1}{5n+5}}\\
&=\color{red}{\frac15\sum \limits_{n=0}^\infty \frac{1}{n+1}}\\
&=\color{red}{\frac15\sum \limits_{n=1}^\infty \frac{1}{n}}\\
&\implies \text{harmonic series} \implies divergent
\end{align}
$\sum \limits_{n=0}^\infty \dfrac{(-1)^{n+1}}{5n+1}$ is convergent but not absolute convergent. $_\blacksquare$
 [/INST] You first part is fine for the second that step is wrong
$$\sum \limits_{n=0}^\infty \frac{1}{5n+1}
\color{red}{=\sum \limits_{n=1}^\infty \frac{1}{5n}}$$
we can simply refer directly to limit comparison test with $\sum \frac 1n$ and conclude for divergence indeed
$$\frac{\frac{1}{5n+1}}{\frac1n}=\frac{n}{5n+1}\to \frac15$$
or as an alternative
$$\sum \limits_{n=0}^\infty \frac{1}{5n+1}\ge \sum \limits_{n=0}^\infty \frac{1}{5n+5}=\frac15\sum \limits_{n=0}^\infty \frac{1}{n+1}=\frac15\sum \limits_{n=1}^\infty \frac{1}{n}$$
</s>