Power Series Convergence Assistance

[theCalc]

The power series

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

converges to what number?

So far, I've tried using the Ratio Test and the limit as n approaches infinity equals $0$. Also since $L<1$, the power series converges by the Ratio Test.

 

[MarkFL]

Maclaurin tells us:

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

You can write the given series in terms of the above. :)

 

[theCalc]

I'm a bit lost and confused as to how I would manipulate the series, could you explain what steps would I need to take? Thanks.

 

[MarkFL]

Well, I think the first thing I would do is write:

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

Next, let's write:

$$S=\sum_{n=2}^{\infty}\left(\frac{(-1)^n(n-1)}{n!}\right)=\sum_{n=2}^{\infty}\left(\frac{(-1)^nn}{n!}\right)-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)$$

$$S=\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{(n-1)!}\right)-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)$$

$$S=\sum_{n=1}^{\infty}\left(\frac{(-1)^{n+1}}{n!}\right)-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)$$

$$S=1-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)-\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)$$

$$S=1-2\sum_{n=2}^{\infty}\left(\frac{(-1)^n}{n!}\right)=\,?$$