$\displaystyle\sum\limits_{n=1}^\infty \left(\frac {x^2}{x^2+4}\right)^n$
I saw this question asked before, but I wasn't sure about the last answer provided for this question, which I'll try to cite here:
https://math.stackexchange.com/q/1305091
I wasn't sure because I thought the series was of the form:
$\displaystyle\sum\limits_{n=1}^\infty ar^{n-1} = a + ar + ar^2 + ar^3+\dots$
And, that I would need to use:
$a[ 1 + r + r^2 + r^3 + \dots]$
to find the first term, which is $\frac{1}{5}$.
I then found the limiting value of the series to be $\left(\frac{\sqrt x + 4}{20}\right)$.
Does this work or am I off?
Best Answer
@user beat me to this, but I'll post anyway. By the ratio test, $\lim\limits_{n\to\infty} \biggr\rvert\dfrac{a_{n+1}}{a_n}\biggr\lvert =\dfrac{x^2}{x^2+4}<1$ for any $x$, and so the limit holds for all $x$. The series is essentially a geometric series with $|r|<1$. Its sum is given by $\dfrac{a}{1-r}$, where $a$ is the first term.