009C Sample Midterm 3, Problem 5

Find the radius of convergence and the interval of convergence of the series.

(a) (6 points)      ${\displaystyle \sum _{n=0}^{\infty }}{\frac {(-1)^{n}x^{n}}{n+1}}.$

(b) (6 points)      ${\displaystyle \sum _{n=0}^{\infty }}{\frac {(x+1)^{n}}{n^{2}}}.$

When we do, the interval will be $(c-r,c+r)$. However, the boundary values for $x$, $c-r$ and $c+r$ must be tested individually for convergence. Many times, one boundary value will produce an alternating, convergent series while the other will produce a divergent, non-alternating series. As a result, intervals of convergence may open, half-open or closed.

Foundations:
When we are asked to find the radius of convergence, we are given a series where
$a_{n}=f(x-c)\cdot g(n)$
where $f$ and $g$ are functions of $x$ and $n$ respectively, and $c$ is a constant (frequently zero). We need to find a bound (radius) on $|x-c|$ such that whenever $|x-c|<r$, the ratio test
$\left|{\frac {a_{n+1}}{a_{n}}}\right|<1.$

 Solution:

(a):
We need to choose a radius $r$ such that whenever $|x|<r$,
${\begin{array}{rcl}\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=&\lim _{n\rightarrow \infty }\left|{\frac {\frac {(-1)^{n+1}x^{n+1}}{n+2}}{\frac {(-1)^{n}x^{n}}{n+1}}}\right|\\\\&=&\lim _{n\rightarrow \infty }\left|x\cdot {\frac {n+1}{n+2}}\right|\\\\&=&|x|\cdot \lim _{n\rightarrow \infty }\left|{\frac {n+1}{n+2}}\right|\\\\&=&|x|<1.\end{array}}$
In this case, the radius is 1, and the interval will be centered at 0. We then need to take a look at the boundary points. If $x=1,$ then
$a_{n}\,=\,{\frac {(-1)^{n}x^{n}}{n+1}}\,=\,{\frac {(-1)^{n}}{n+1}},$
so the series is an alternating harmonic series which converges. On the other hand, if $x=-1,$ then
$a_{n}\,=\,{\frac {(-1)^{n}x^{n}}{n+1}}\,=\,{\frac {(-1)^{n}(-1)^{n}}{n+1}}\,=\,{\frac {1}{n+1}},$
a standard harmonic series which does not converge. Thus, the interval of convergence is $(-1,1]$.

(b):
We need to choose a radius $r$ such that whenever $|x|<r$,
${\begin{array}{rcl}\lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=&\lim _{n\rightarrow \infty }\left|{\frac {\frac {(x+1)^{n+1}}{(n+1)^{2}}}{\frac {(x+1)^{n}}{n^{2}}}}\right|\\\\&=&\lim _{n\rightarrow \infty }\left|(x+1)\cdot \left({\frac {n}{n+1}}\right)^{2}\right|\\\\&=&|x+1|\cdot \lim _{n\rightarrow \infty }\left({\frac {n}{n+1}}\right)^{2}\\\\&=&|x+1|<1.\end{array}}$
In this case, the radius is 1, and the interval will be centered at $x=-1$, or when $x+1=0$. We then need to take a look at the boundary points. If $x=-2$ or $x=0$, then
$\left|a_{n}\right|\,=\,\left|{\frac {(x+1)^{n}}{n^{2}}}\right|\,=\,{\frac {1}{n^{2}}},$
which defines a p-series with $p=2$. Thus, the series defined by each boundary point is absolutely convergent (and therefore convergent), and the interval of convergence is $[-2,0]$.

Final Answer:
For (a), the radius is 1 and the interval of convergence is $(-1,1]$.
For (b), the radius is also 1, but the interval of convergence is $[-2,0]$.

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