show that $ 4 = \sum_{n=1}^\infty (-2)^{n+1}\frac{n+2}{n!} $

Question

I need to show that $$ 4 = \sum_{n=1}^\infty (-2)^{n+1}\frac{n+2}{n!} $$

by considering $$ \frac d{dx}(x^2e^{-x})$$

I found that $ \frac d{dx}(x^2e^{-x}) = 2xe^{−x}−x^2e^{−x}$

What would be the next step to show the equality?

score 4 · Answer 1 · answered Jan 21 '15 at 10:23

4

$$x^2e^{-x}=x^2\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}=\sum_{n=0}^\infty\frac{(-1)^nx^{n+2}}{n!}\implies$$

$$2xe^{-x}-x^2e^{-x}=\sum_{n=0}^\infty (-1)^n\frac{(n+2)x^{n+1}}{n!}$$

...and the above is true for all $\;x\in\Bbb R\;$ .

answered Jan 21 '15 at 10:23

Timbuc

To arrive at the answer if I multiply both sides by 2 and put $x=1$ then I am getting answer as $\frac{4}{e}\neq 4$ .Please help – Learnmore Jan 21 '15 at 10:38
@learnmore 2*(-1)^n is not (-2)^n – TenaliRaman Jan 21 '15 at 10:53
@learnmore Why would you do such a thing?? Read Tenali's comment above: $$(-1)^nx^{n+1}=x\cdot(-x)^n$$ – Timbuc Jan 21 '15 at 11:07
so what's the way out @Timbuc – Learnmore Jan 21 '15 at 11:14
@learnmore Substitute$;x=2;$ and note your series begins with $;n=1;$ – Timbuc Jan 21 '15 at 11:21
@Timbuc
$$2xe^{-x}-x^2e^{-x}=\sum_{n=0}^\infty (-1)^n\frac{(n+2)x^{n+1}}{n!} = \sum_{n=0}^\infty x(-x)^n\frac{(n+2)}{n!}$$ and x can be cancelled. But when I plug x=2 the left side becomes 0
– Togepi Jan 21 '15 at 18:46
@Togepi Exactly!! It becomes zero, and your series begins with $;n=1;$ and the first summand is $;4;$ ....do it! – Timbuc Jan 21 '15 at 19:45
@Timbuc I got it thanks alot!! – Togepi Jan 21 '15 at 20:01

score 0 · Answer 2 · answered Jan 21 '15 at 10:26

0

Hint: It seems that the equality $\frac{n}{n!} = \frac{1}{(n-1)!}$ for $n\ge 1$ and a shift in the summation index can be used.

answered Jan 21 '15 at 10:26

orangeskid

2 Answers2