# Convergence of $\sum_{n=0}^\infty(-1)^n\frac{4^{n-2}(x-2)}{(n-2)!}$

What theorem should I use to show that $$\sum_{n=0}^\infty(-1)^n\frac{4^{n-2}(x-2)}{(n-2)!}$$ is convergent no matter what value $x$ takes?

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Did you tried using the ratio test? –  Integral Jul 11 '12 at 1:51
Is $(x-2)$ supposed to be $(x-2)^n$? –  Pedro Tamaroff Jul 11 '12 at 1:52
Is just $(x-2)$ –  hinafu Jul 11 '12 at 1:54
Then why should it's convergence depend on $x$? The convergence in such case depends only on the convergence of the series. And, careful since you're getting terms with $(-2)!$ and $(-1)!$ in them. Could you double check what you wrote is correct? –  Pedro Tamaroff Jul 11 '12 at 1:56
That's what I'm trying to prove, that $x$ works as a constant, and yes, it is $(n-2)!$ but I can change it to $\Gamma(n-1)$ –  hinafu Jul 11 '12 at 1:59

Note that while $(-2)!$ and $(-1)!$ are divergent, $1/(-2)! = 1/(-1)! = 0$. Effectively the sum starts at $n=2$.

Let $\displaystyle a_n = (-1)^n\frac{4^{n-2}(x-2)}{(n-2)!}$. Then $$\left|\frac{a_{n+1}}{a_n}\right| = \frac{4}{n-1},$$ so the sum $\sum_{n=2}^\infty |a_n|$ converges by the ratio test. Now use the fact that absolutely convergent series are convergent. The sum converges to $(x-2)/e^4$, independent of the value of $x$.

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Your teacher must have given you this.

$$\sum_{n=2}^\infty(-1)^n\frac{4^{n-2}(x-2)^{n-2}}{(n-2)!}=\sum_{n=0}^\infty(-1)^{n+2}\frac{4^{n}(x-2)^n}{n!}=\sum_{n=0}^\infty(-1)^{n}\frac{(4(x-2))^n}{n!}=e^{-4(x-2)}$$ for all $x \in \mathbb{R}$(by definition) which proves that regardless of the value of $x$ the series converges.

or may be this,

$$\sum_{n=2}^\infty(-1)^n\frac{4^{n-2}(x-2)^{n}}{(n-2)!}=(x-2)^2\sum_{n=0}^\infty(-1)^{n}\frac{(4(x-2))^n}{n!}=(x-2)^2e^{-4(x-2)}$$ for all $x \in \mathbb{R}$ (by definition)

or may be this,

$$\sum_{n=2}^\infty(-1)^n\frac{4^{n-2}(x-2)}{(n-2)!}=(x-2)\sum_{n=0}^\infty(-1)^{n}\frac{4^n}{n!}=(x-2)e^{-4}$$ for all $x \in \mathbb{R}$ (by definition)

each of which are obviously convergent for all $x$.

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Note that $(-1)^n = (-1)^{n-2}$. Hence, $$\sum_{n=0}^\infty(-1)^n\frac{4^{n-2}(x-2)}{(n-2)!} = (x-2) \sum_{n=0}^\infty\frac{(-4)^{n-2}}{(n-2)!} = (x-2) \left(\sum_{n=0}^\infty\frac{(-4)^{n}}{n!} \right)$$ where we have interpreted $\dfrac1{(-1)!} = 0 = \dfrac1{(-2)!}$.

This is a reasonable interpretation since $\dfrac1{\Gamma(0)} = 0 = \dfrac1{\Gamma(-1)}$.

Now recall that $$\sum_{n=0}^\infty\frac{y^{n}}{n!} = \exp(y).$$ Can you now conclude that the series converges no matter what value $x$ takes?

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thanks a lot! this should convince my teacher –  hinafu Jul 11 '12 at 2:38

Ratio (D'alembert's) test, as proposed by Integral: $$a_n:=(-1)^n\frac{4^{n-2}}{(n-2)!}\Longrightarrow \left|\frac{a_{n+1}}{a_n}\right|=\frac{4^{n-1}}{(n-1)!}\frac{(n-2)!}{4^{n-2}}=\frac{4}{n-1}\xrightarrow [n\to\infty]{} 0$$

so that the convergence radius is $\,\infty\,$.

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By "theorem", your teacher probably just means that $(x-2)$ can be moved outside of the series (distributive property over infinite sums), so that its convergence doesn't depend on $x$. Proving the convergence of the series itself would require the series to actually make sense, which it currently doesn't. Though if you nudge the index forward so that all values are defined it will certainly converge, as you can see by applying the ratio test, which may also be what your teacher was referring to.

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