power series

[kezman]

I have to find the values of x$\displaystyle \in R$ for which the series converge

$\displaystyle \sum\limits_{n = 1}^\infty {n!(x + 1)^n }$

I having difficulties to find any solution.

 

[skeeter]

I having difficulties to find any solution.

so am I ... the series diverges for all values of x.

 

[kezman]

I having difficulties to find any solution.

so am I ... the series diverges for all values of x.

Thats what I thought.

Im having similar problems with this series:

$\displaystyle \sum\limits_{n = 1}^\infty {2^n } \sin \frac{x}{{3^n }}$

 

[pka]

Actually $\displaystyle \sum\limits_{n = 0}^\infty {n!(x + 1)^n}$ converges for x=-1, with radius 0. [one point convergence].

 

[kezman]

thats right...

Is there another way of knowing that without repalcing x with -1?

 

[skeeter]

well ... if what pka sez is true (and he's very reliable), then

\(\displaystyle \L\Sigma 2^n sin(\frac{x}{3^n})\) converges for all values of x that are of the form $\displaystyle k \pi$, where k is an integer.

but ... I think this series converges for all x.

$\displaystyle lim_{n\rightarrow \infty} \frac{sin(\frac{x}{3^n})}{\frac{x}{3^n}} \, = \, 1$

so wouldn't this series have the same interval of convergence as
\(\displaystyle \L\Sigma \frac{2^n}{3^n} x \, = \, \Sigma (\frac{2}{3})^n x\) ???

 

[pka]

More to the point, $\displaystyle x > 0\quad \Rightarrow \quad \sin (x) < x.$. Be careful for x<0.
However, what about absolute convergence?

 

[kezman]

If I start from the point of the geometric series.
For the two methods I have these results...

\(\displaystyle \begin{array}{l}
\sum\limits_{n = 0}^\infty {\sqrt[n]{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\left| x \right|}}} < 1 \Rightarrow \sqrt[n]{{\left| x \right|}} < \frac{3}{2} \\
\sum\limits_{n = 0}^\infty {\frac{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\frac{2}{3}\left| x \right|}}{{\left| {\left( {\frac{2}{3}} \right)^n } \right|\left| x \right|}}} < 1 \Rightarrow \frac{2}{3} < 1 \\
\end{array}\)

 

[pka]

The point is: use the basic comparison test.

 

[kezman]

The point is: use the basic comparison test.

$\displaystyle {\frac{{{\rm 2}^{\rm n} \sin \frac{x}{{3^n }}}}{{2^n \frac{x}{{3^n }}}}} = 1$
so if Bn converges An converges
and

$\displaystyle {2^n \frac{x}{{3^n }}}$

converges for all x

 

[pka]

NO! NOT AT ALL THAT. Here is how it works.

If $\displaystyle \sum\limits_n {\left| {a_n } \right|}$ converges then $\displaystyle \sum\limits_n {a_n }$ converges. (ABSOLUTE CONVERGENCE)

If $\displaystyle \forall n,\;0 \le a_n \le b_n$ and $\displaystyle \sum\limits_n {b_n }$ converges then $\displaystyle \sum\limits_n {a_n }$ converges. (BASIC COMPARSION)


Recall that $\displaystyle \left| {\sin (x)} \right| \le \left| x \right|$
so that $\displaystyle \sum\limits_n {\left| {2^n \sin \left( {\frac{x}{{3^n }}} \right)} \right|} \le \sum\limits_n {\left| {2^n \left( {\frac{x}{{3^n }}} \right)} \right| \le|x|\sum\limits_n {\left| {\left( {\frac{2}{3}} \right)^n } \right|} }$ the last series converges for any x.