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. $_\blacksquare$

  • 1
    $\begingroup$ "We still need to show..." . No, you already did (almost) that above since it is only the absolute value of the general term. What your proof is lacking in order to use Leibniz criterion is to show that the absolute value of the general term is monotone descending ... Also, we have that $$\sum_{n=0}^\infty\frac1{5n+1}\neq\sum_{n=1}^\infty\frac1{5n}$$ Not even close. Some more care is needed here. $\endgroup$ – DonAntonio Oct 29 '18 at 22:19
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    $\begingroup$ @gimusi In fact he "noted" wrongly that the actual general term is monotonic, which of course it isn't as is an alternating one. He should take away the $\;(-1)^{n}\;$ (absolute value) and then proceed to show $\endgroup$ – DonAntonio Oct 29 '18 at 22:46
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    $\begingroup$ @Doesbaddel Yes now it is fine. Note that this step $\frac{1}{(5n+1)(5n+2)}\geq 0 1\implies \frac{1}{2 + 15 n + 25 n^2}\geq 0$ is not necessary. $\endgroup$ – user Oct 29 '18 at 23:28
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    $\begingroup$ @Doesbaddel A proof is fine when we use the minimum symbol and step to obtain the result. Therefor you should try always to avoin unnecessary steps. But it is fine! Bye $\endgroup$ – user Oct 29 '18 at 23:30
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    $\begingroup$ @Doesbaddel Your proof of monotonicity is incorrect: if $\;a_n=\frac1{5n+1}\;$ , then$$\;a_{n+1}=\frac1{5(n+1)+1}=\frac1{5n+6}\;$$ These numerous little mistakes are mostly due to lack of proper care when working. Don't worry, with constant work they're due to disappear gradually. $\endgroup$ – DonAntonio Oct 29 '18 at 23:31

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

$$\sum \limits_{n=0}^\infty \frac{1}{5n+1}\ge \sum \limits_{n=0}^\infty \frac{1}{5n+5}=\frac15\sum \limits_{n=0}^\infty \frac{1}{n+1}=\frac15\sum \limits_{n=1}^\infty \frac{1}{n}$$

  • $\begingroup$ I completely overlooked that equation and started doing Index-Shifts like you would do it for $\sum \limits_{n=0}^\infty \frac{1}{n+1} = \sum \limits_{n=1}^\infty \frac{1}{n}$ Thank you for pointing that error out. $\endgroup$ – Doesbaddel Oct 29 '18 at 22:32
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    $\begingroup$ @Doesbaddel Also consider the limit comparison test, here it is not necessary but in many other cases it can be very useful. $\endgroup$ – user Oct 29 '18 at 22:33

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