0
$\begingroup$

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 $a_n$?

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ So term-by-term comparison is way to go in series where stuff like $1! + 2! + \cdots + n!$ or $1 + 2 + \cdots + n$ occur? $\endgroup$ – Accelerate to the Infinity Jun 6 '17 at 15:50
  • 1
    $\begingroup$ @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 $$\sum_{n=1}^{\infty}\frac{(-1)^n}{(n+1)!}\left[\sum_{k=1}^{n} k!\right]$$) $\endgroup$ – user12345 Jun 6 '17 at 15:54
2
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.