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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [-2,0]} .

Final Answer:
For (a), the radius is 1 and the interval of convergence is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (-1,1]} .
For (b), the radius is also 1, but the interval of convergence is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [-2,0]} .

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