Leibniz test for convergence of non alternating series I am aware that one can use the comparison test and the integral test to show that the series $$\sum_{n=1}^{\infty}\frac{1}{n(n+3)}$$ converges. Is it possible to use the Leibniz test to show that the series converges?
 A: By telescoping  the series, we have
$$
\begin{align}
\sum_{n=1}^N\frac1{n(n+3)}
&=\frac13\sum_{n=1}^N\left(\frac1n-\frac1{n+3}\right)\\
&=\frac13\sum_{n=1}^N\frac1n-\frac13\sum_{n=4}^{N+3}\frac1n\\
&=\frac13\left(\sum_{n=1}^3\frac1n+\sum_{n=4}^N\frac1n\right)-\frac13\left(\sum_{n=4}^N\frac1n+\sum_{n=N+1}^{N+3}\frac1n\right)\\[3pt]
&=\frac13\left(1+\frac12+\frac13\right)-\frac13\left(\frac1{N+1}+\frac1{N+2}+\frac1{N+3}\right)
\end{align}
$$
Now let $N\to\infty$.
A: Simply notice that
$$\frac1{n(n+3)}\le \frac1{n(n+1)}=\frac1{n}-\frac1{n+1}$$
and by telescoping
$$\sum_{n=1}^\infty \frac1{n}-\frac1{n+1}=1$$
so the given series is convergent by comparison.
A: Your sequence $$\sum\frac1{n(n+3)}=\frac14+\frac1{10}+\frac{1}{18}+\frac1{28}+\frac1{40}+\cdots$$
$$=\sum\left(\frac1n-\frac{n+2}{n(n+3)}\right)=1-\frac{3}{4}+\frac12-\frac4{10}+\frac13-\frac{5}{18}+\frac14-\frac{6}{28}+\frac15-\frac{7}{40}\pm\cdots$$  
$$=\sum\left(\frac{n+4}{n(n+3)}-\frac1n\right)=\frac{5}{4}-1+\frac6{10}-\frac12+\frac7{18}-\frac13+\frac{8}{28}+\frac14+\frac{9}{40}-\frac15\pm\cdots$$
Those are the two nice cases of $\sum\left(\frac{n+A+1}{n(n+3)}-\frac{n+A}{n(n+3)}\right)$
Another variation is  $=\sum\left(\frac{2n+A+1}{n(n+3)}-\frac{2n+A}{n(n+3)}\right).$ For $A=3$ this gives $$\frac{5}{4}-\frac{4}{4}+\frac{7}{10}-\frac{6}{10}+\frac{9}{18}-\frac{8}{18}+\frac{11}{28}-\frac{10}{28}+\frac{13}{40}-\frac{12}{40}\pm \cdots  $$
This has the nice feature that , ignoring signs, each term is close to half way between those on either side.

In some sense one can always do this. Given a convergent positive series $\sum a_i$ we wish to rewrite it as an alternating series $$a_1+a_2+a_3\cdots=b_1-b_2+b_3-b_4+b_5-b_6\pm\cdots$$ with 


*

*$ b_{2n-1}-b_{2n}=a_n$ 

*$b_{2n} \gt b_{2n+1}$

*all the $b_k$ are positive with $\lim b_k=0$


The series would instead be called telescoping if we changed the second condition to $b_{2n}=b_{2n+1}.$ Then we see that everything is forced. $b_{2n-2}=b_{2n-1}=\sum_{k}^{\infty}a_n.$ The catch is that we might not recognize the $b_k$ as anything nice. 
For the given series $b_{2n-1}=\frac{11}{18},\frac{13}{36},\frac{47}{180},\frac{37}{180},\cdots$ 
It took me a long time to realize that this is pretty close to $\frac12,\frac13,\frac14,\frac15\cdots.$  $$
b_{2n-1}=\frac{1}{n+1}+\frac{2}{3n(n+1)(n+2)}=\frac{3n^2+6n+2}{3n(n+1)(n+2)}=\frac{n+1}{n(n+2)}-\frac{1}{3n(n+1)(n+2)}. $$ You can prove that after the fact, but that  isn't how I found it.
Evidently for an honest alternating series we must have $b_{2n-1} \gt \sum_n^{\infty}a_k.$ Then $b_{2n}=b_{2n-1}-a_n$ and we will always have a little room to choose the next term with $b_{2n} \gt b_{2n+1} \gt \sum_{n+1}^{\infty}a_k.$
Again it may be difficult to make this nice.
