3
$\begingroup$

I have to evaluate the following sum:

$$\sum_{n=0}^{\infty} \frac{n^{2}-5n+2}{n!}$$

Could you recommend some guide for solving this? I have not found any information that I could use for this type of exercise.

2 Answers 2

6

$$\frac{n^2-5n+2}{n!}=\frac{n^2}{n!}-5\frac n{n!}+2\frac1{n!}$$

$$\frac{n^2}{n!}=\frac{n(n-1)+n}{n!}=\frac1{(n-2)!}+\frac1{(n-1)!}$$

$$\frac n{n!}=\frac1{(n-1)!}$$

where $\frac1{(-2)!}=\frac1{(-1)!}=0$ so that we have

$$S=\sum_{n=0}^\infty\frac1{(n-2)!}-\frac4{(n-1)!}+\frac2{n!}=e-4e+2e=-e$$

  • 0
    I'm sorry but you can explain the steps is that I do not understand very well and if I compare it with this other exercise $ \sum_{n=1}^{\infty}\frac {(n-1) 3} {n!} $, The resolution is very different. – 2017-01-21
  • 0
    @eslop Ok. Which steps do you need explanation? – 2017-01-21
  • 0
    in the second step, what happen with $-5\frac{n}{n!}+2\frac{1}{n!}$? – 2017-01-21
  • 0
    @eslop The $-5\frac n{n!}$ goes to $-5\frac1{(n-1)!}$ and I did nothing with the $\frac2{n!}$ – 2017-01-21
  • 0
    And the last question is the rule by which n! Is divided into (n-1) ... (n-2) ...? In the other exercise \frac{(n-1)^3}{n!} is split in n! (n-1)! (n-2)! (n-3)! – 2017-01-21
  • 0
    @eslop Ugh, I wouldn't do it that way. Just my preference... – 2017-01-21
  • 0
    and if the $\sum_{n=1}^{\infty}$ or $\sum_{n=2}^{\infty}$ is equal? – 2017-01-22
  • 0
    @eslop Yes, as it follows from $\frac1{(-2)!}=\frac1{(-1)!}=0$, so those terms become negligible. – 2017-01-22
  • 0
    Using this same process in this exercise $\sum_{n = 1}^{\infty} \frac {(n-1) 3} {n!}$ I get e, but my teacher came out e + 1 could you tell me what the difference? Thanks – 2017-01-22
  • 0
    @eslop Is that supposed to say $(n-1)^3$? – 2017-01-22
  • 0
    yes is $(n-1)^{3}$ sorry – 2017-01-22
  • 0
    By the way, it's probably too lengthy, so you probably should post as another question... or let's go to [my chat room](http://chat.stackexchange.com/rooms/51337/simply-beautiful-arts-room) – 2017-01-22
3

Let us make the problem more general considering $$F(x)=\sum_{n=0}^{\infty} \frac{an^{2}+bn+c}{n!}x^n$$ First, write $$an^2+bn+c=a(n(n-1)+n)+bn+c=an(n-1)+(a+b)n+c$$ $$F(x)=a\sum_{n=0}^{\infty} \frac{n(n-1)}{n!}x^n+(a+b)\sum_{n=0}^{\infty} \frac{n}{n!}x^n+c\sum_{n=0}^{\infty} \frac{x^n}{n!}$$ $$F(x)=ax^2\sum_{n=0}^{\infty} \frac{n(n-1)}{n!}x^{n-2}+(a+b)x\sum_{n=0}^{\infty} \frac{n}{n!}x^{n-1}+c\sum_{n=0}^{\infty} \frac{x^n}{n!}$$ $$F(x)=ax^2\left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right)''+(a+b)x\left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right)'+c \left(\sum_{n=0}^{\infty} \frac{x^n}{n!}\right)$$ $$F(x)=ax^2e^x+(a+b)xe^x+c e^x$$ $$F(1)=(2a+b+c)\,e$$

If you had faced the problem of $$G(x)=\sum_{n=0}^{\infty} \frac{an^{3}+bn^2+cn+d}{n!}x^n$$ writing $n^3=n(n-1)(n-2)+3n(n-1)+n$ and applying the same approach, you would obtain $$G(1)= (5 a+2 b+c+d)\,e$$ and so on.

  • 0
    One may also note, as a third method, that $$\sum_{n=0}^\infty\frac{n^k}{n!}=B_ke$$where we use the Bell numbers. – 2017-01-22