Determine whether the sum

enter image description here

is divergent, absolute convergent or conditionally convergent.

First, I tried to solve this by hand. I tried to find a convergent majorant series for the series in question.

$$\bigg|\cos(n^2+1) \frac{2+n^2}{1+n!} \bigg| \leq\frac{n^2}{n!}$$

It turns out that $\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$, according to Maple that is. This result shows that the sum in the question is absolute convergent.

I also tried to use Maple to double check my answer, and it gave me this.

enter image description here

When Maple gives this sort of output, it usually means that the sum is not convergent. But didn't I just show that it was, or have I made a mistake?

Can a "Maple man" tell me why Maple gives this result, and perhaps how I can avoid it?

  • 1
    $\begingroup$ The series does converge, but only a maple person can tell you why maple doesn't recognize it as such. Can you use maple to make a table of partial sums for some large values of n? $\endgroup$ Dec 8, 2020 at 15:17
  • 3
    $\begingroup$ It will be related to the formal=false parameter, without it I get it is approximately -1.379777889 $\endgroup$
    – Sil
    Dec 8, 2020 at 15:19

2 Answers 2


The sum converges absolutly. Your proof is correct (exept that inequlity $\frac{n^2+2}{1+n!} \leq \frac{n^2}{n!}$ is not true, but the last fraction may be replaced by $\frac{2+n^2}{n!}$, which give us convergent series).


As it was already mentioned by Timur B., your proof is correct except a moment $\bigg|\cos(n^2+1) \frac{2+n^2}{1+n!} \bigg| \leq\frac{2+n^2}{n!}$ instead of $\frac{n^2}{n!}$.

I will add that if we see that we shouldn't trust to Maple, so it's better not to use Maple to find $\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$ in order to finish the proof) Although this time the statement is correct and indeed $\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$,

it's easier to notice that for $n > 100$ we have $\bigg|\cos(n^2+1) \frac{2+n^2}{1+n!} \bigg| \leq\frac{2+n^2}{n!} \le \frac{2n^2}{100 n(n-1)(n-2)(n-3)} <$$ $$ < \frac{2n^2}{100n \cdot \frac{n}2 \cdot \frac{n}2 \cdot \frac{n}2} \le \frac{1}{n^2} \le \frac{1}{n(n-1) } = \frac1{n-1} - \frac{1}n,$ and the series $\sum_{n \ge 100} ( \frac1{n-1} - \frac{1}n ) = (\frac{1}{99} - \frac{1}{100}) + (\frac{1}{100} - \frac{1}{101}) + \ldots ... $ converges (to $\frac{1}{99}$) obviously.


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