Series Convergence: Test for x Values

[karush]

\(\displaystyle (x-1)-\frac{(x-1)^2}{2!}+\frac{(x-1)^3}{3!}-\frac{(x-1)^4}{4!}+ ∙ ∙ ∙\)

well this looks like an alternating-series, the question is: at what value(s) of x does this
converge.

one observation is that if x=0 then all terms are 0 so there is no convergence, also I presume you can rewrite this as
$\frac{(x-1)^n}{n!}$ and then test for values of $x$

 

[chisigma]

\(\displaystyle (x-1)-\frac{(x-1)^2}{2!}+\frac{(x-1)^3}{3!}-\frac{(x-1)^4}{4!}+ ∙ ∙ ∙\)

well this looks like an alternating-series, the question is: at value(s) of x does this
converge.

one observation is that is if x=0 then all terms are 0 so there is no convergence, also I presume you can rewrite this as
$\frac{(x-1)^n}{n!}$ and then test is for values of $x$

Remembering the well known series...

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

... that converges for any real or complex value of z, it is easy to observe that is...

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

... and the series (2) converges for any value of x...

Kind regards

$\chi$ $\sigma$

 

[Prove It]

\(\displaystyle (x-1)-\frac{(x-1)^2}{2!}+\frac{(x-1)^3}{3!}-\frac{(x-1)^4}{4!}+ ∙ ∙ ∙\)

well this looks like an alternating-series, the question is: at what value(s) of x does this
converge.

one observation is that if x=0 then all terms are 0 so there is no convergence, also I presume you can rewrite this as
$\frac{(x-1)^n}{n!}$ and then test for values of $x$

Even though recognising this series as an exponential function is the quickest method, if you didn't realize that, then you would have to apply the ratio test.

The ratio test states that for a series $\displaystyle \begin{align*} \sum{ a_n } \end{align*}$, the series will be convergent if $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \end{align*}$. But in your case, you have a function of x, which gives a different series for every x that is put in. So what you have to do is to evaluate $\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \end{align*}$ in terms of x, then set this expression less than 1 and solve for x. This will tell you nearly all the values of x for which your function gives a convergent series (you need to check the points where this limit is equal to 1 separately, as the ratio test is inconclusive there...)