# Proving monotonicity for Leibniz cirterion

I'm given series $\sum_{n = 1}^{+\infty} \frac{(-1)^{n}}{(n+1)!}\left(1 + 2! + \cdots + n!\right)$ and I have to find whether it is convergent.

Testing for absolute convergence, we have $a_n = \frac{1}{(n+1)!} + \frac{2}{(n+1)!} + \cdots + \frac{(n-1)!}{(n+1)!} + \frac{n!}{(n+1)!}$ and since last term is $\frac{n!}{(n+1)!} = \frac{1}{n+1}$ series diverge in comparison with harmonic series and hence can only be conditionally convergent, which I will try to prove from Leibniz criterion.

Now, I have to show, that $a_n$-th term is monotonically decreasing and $\lim a_n = 0$.

Treating $a_n$ as $\frac{a_n}{b_n} = \frac{1! + 2! + \cdots + n!}{(n+1)!}$ I can use Stolz-Cesàro theorem ($\lim \frac{a_n}{b_n} = \lim\frac{a_{n+1} - a_n}{b_{n+1} - b_n}$) since $b_n$ is monotonically increasing and $\lim b_n = +\infty$. Then $$\lim \frac{a_n}{b_n} = \lim\frac{a_{n+1} - a_n}{b_{n+1} - b_n} = \lim\frac{(n+1)!}{(n+2)! - (n+1)!} = \lim \frac{1}{n+2}\frac{1}{1 - \frac{1}{n+2}} = 0.$$

But how to prove monotonicity? I've tried $\frac{a_{n+1}}{a_n}$ but it didn't get me anywhere. What are some ways to show monotonicity of sequences like $a_n$?

Pairwise comparison gives us: $$a_{n+1}=\dfrac{1!+2!+\ldots+n!+(n+1)!}{(n+2)!}$$ $$=\underbrace{\dfrac{1}{(n+2)!}}_{<\frac{1}{(n+1)!}}+\underbrace{\dfrac{2}{(n+2)!}}_{<\frac{2}{(n+1)!}}+\ldots+\underbrace{\dfrac{1}{(n+1)(n+2)}}_{<\frac{1}{n(n+1)}}+\underbrace{\dfrac{1}{n+2}}_{<\frac{1}{n+1}}<a_{n}$$ Therefore, $a_{n+1}<a_{n}$ and it's decreasing.

• So term-by-term comparison is way to go in series where stuff like $1! + 2! + \cdots + n!$ or $1 + 2 + \cdots + n$ occur? – Accelerate to the Infinity Jun 6 '17 at 15:50
• @AcceleratetotheInfinity If it works, it works, but yeah, in this case since we're just trying to show an inequality and each term of $a_{n+1}$ is smaller than its "corresponding" term in $a_{n}$, then it works very well (assuming I did not make a mistake). I've used it several times with some nested sums of this form (this one is $$\sum_{n=1}^{\infty}\frac{(-1)^n}{(n+1)!}\left[\sum_{k=1}^{n} k!\right]$$) – user12345 Jun 6 '17 at 15:54

Consider $a_{n-1}-a_{n}$. This is equal to

$$\frac{1}{n!}-\frac{1}{(n+1)!} + 2(\frac{1}{n!}-\frac{1}{(n+1)!}) + \ldots + n(\frac{1}{n!}-\frac{1}{(n+1)!}) - \frac{n}{(n+1)!}$$

which simplifies to

$$\frac{n(1+2+\ldots +n)}{(n+1)!} - \frac{n}{(n+1)!}$$

which is clearly greater than 0 for all $n>1$ and so $a_n>a_{n-1}$ for all $n>1$, which gives the result