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Find values of $x$ where

$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(x-2)^n}{n2^n}$$

is convergent.

So I think ...

$$c_n = \frac{(-1)^{n+1}}{n2^n}$$

But is this right, $n$ shouldn't be in $c_n$ right? I am very lost when it comes to power series ...

My other tries I believe are wrong:

Try 1

Since this is a alternating series, I tried $\lim_{n\to\infty} |a_n| = 0$

$$\lim_{n\to\infty} \frac{(x-2)^n}{n2^n} = 0$$

It appears the eqality hold only when $x=2$ but it is not decreasing then.

Try 2

The above where I try to find $c_n$ then the radius of convergence, but I think my $c_n$ is wrong?

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hint: set $t=\frac{x-2}2$ you should recognize a famous Taylor series. – Raymond Manzoni Mar 18 '12 at 12:18
up vote 1 down vote accepted

Your $c_n$ are fine. A power series has the form $$ \sum_{n=1}^\infty c_n(x-a)^n $$ where the coefficients $c_n$ are scalars that do not depend on $x$, they can depend on $n$.

Your series is $$ \sum_{n=1}^\infty {(-1)^{n+1}\over n 2^n}(x-2)^n $$ Maybe it's illustrative to write out this series more explicitly $$\textstyle {1\over2}(x-2)^1+{-1\over 2\cdot 2^2}(x-2)^2+{1\over 3\cdot 2^3}(x-2)^3+\cdots. $$ The coefficients are the numbers $$ \textstyle {1\over2},\ {-1\over 2\cdot 2^2},\ {1\over 3\cdot 2^3},\ \cdots; $$ or, in general, the coefficients are: $$ c_n={(-1)^{n+1}\over n2^n}. $$ To find the radius of convergence, you can compute (note the absolute values) $$ \lim_{n\rightarrow\infty}{ |c_{n+1}|\over |c_n|} =\lim_{n\rightarrow\infty}{{1\over (n+1) 2^{n+1}}\over{1\over n2^n} } =\lim_{n\rightarrow\infty}{n\over (n+1)2}={1\over2}. $$

The radius of convergence is the reciprocal of the above limit: $1/(1/2)=2$. This tells you that:

$\ \ \ \ $ the series converges whenever $|x-2|<2$


$\ \ \ \ $ the series diverges whenever $|x-2|>2$.

Since $|x-2|<2$ if and only if $0<x<4$, we see that the series converges on the interval $(0,4)$. Outside the interval $[0,4]$, the series diverges.

So we almost have the answer. But, we have not said anything about the endpoints of this interval ($x=0$ and $x=4$). The series may or may not converge at the points $x=0$ and $x=4$. You need to check what happens when $x=0$ and $x=4$ separately.

This is a matter of replacing $x$ by the appropriate value in the series and seeing what you get:

When $x=0$, the series becomes: $$ \sum_{n=1}^\infty {(-1)^{n+1}\over n 2^n}( 0-2)^n =\sum_{n=1}^\infty {(-1)^{2n+1}\over n } =\sum_{n=1}^\infty {-1\over n } $$ which diverges. The series does not converge for $x=0$.

When $x=4$, the series becomes: $$ \sum_{n=1}^\infty {(-1)^{n+1}\over n 2^n}(4- 2)^n= \sum_{n=1}^\infty {(-1)^{n+1}\over n 2^n}( 2)^n =\sum_{n=1}^\infty {(-1)^{n+1}\over n } $$ which is a convergent alternating series. The series converges for $x=4$.

Putting everything together, the interval of convergence for the series is $(0,4]$; that is the series converges if and only if $x$ is in the interval $(0,4]$.

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Thanks for the very detailed explaination, it helped me alot :) – Jiew Meng Mar 18 '12 at 14:10
Err... regarding the convergence at end points isn't the 2nd expression when $x=0$ same as the last for $x=4$, why does 1 diverge while the other converge? Is it a typo? Else how do you get $-1$ from $(-1)^{2n+1}$? – Jiew Meng Mar 18 '12 at 14:25
Oh ic, I think you have a typo, it should be $(-1)^{n+1}$ instead of $(-1)^{2n+1}$. And to answer myself, $2n+1$ always results in odd numbers – Jiew Meng Mar 18 '12 at 14:29
@JiewMeng Yes, thanks for spotting the typo. Cut and paste for the lose :) – David Mitra Mar 18 '12 at 14:32

Let $z=\frac12(x-2)$ and, for every $n\geqslant1$, $x_n=(-1)^{n+1}\frac{(x-2)^n}{n2^n}$. Hence $x_n=(-1)^{n+1}\frac{z^n}n$ and $|x_n|=\frac{|z|^n}n$.

  • If $|z|\gt1$, then $|x_n|\to\infty$ and the series $\sum\limits_nx_n$ diverges. This happens if $x\lt0$ or $x\gt4$.
  • If $|z|\lt1$, then $x_n\to0$ and furthermore $|x_n|\leqslant |z|^n$ for every $n\geqslant1$. Since $0\leqslant |z|\lt1$, this proves the series $\sum\limits_nx_n$ converges absolutely. This happens if $0\lt x\lt4$.
  • If $z=1$, then $x_n=\frac{(-1)^{n+1}}n$ hence the series $\sum\limits_nx_n$ converges (and the value of its sum is $\log2$). This happens if $x=4$.
  • If $z=-1$, then $x_n=-\frac1n$ hence the series $\sum\limits_nx_n$ diverges. This happens if $x=0$.

Finally, the series $\sum\limits_nx_n$ converges if and only if $0\lt x\leqslant4$.

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