Infinite sequence and power series infinite sequence $a_{n}$
where $$\lim_{n\to \infty} |na_{n}|=1101 $$
Find R of convergence of the power series $$\sum_{n=1}^\infty a_{n}x^n$$
Anyone can guide me for this question?
Thank you so much!
 A: we are looking for when $\lim\limits_{n\rightarrow\infty}|x|\left|\frac{a_{n+1}}{a_n}\right|<1$
Now from the first equation, it follows that $\lim\limits_{n\rightarrow\infty}|a_n|=\lim\limits_{n\rightarrow\infty}\frac{1101}{n}$, and so, similarly, $\lim\limits_{n\rightarrow\infty}|a_{n+1}|=\lim\limits_{n\rightarrow\infty}\frac{1101}{n+1}$. Thus,
$$|x|\lim\limits_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_n}\right|=|x|\lim\limits_{n\rightarrow\infty}\frac{1101/(n+1)}{1101/n}=|x|$$
Thus you need $|x|<1$, Or a radius of convergence of $1$.
A: Hint. As $n \to \infty$, you have
$$
|a_n| \sim \frac{1101}n
$$ then, by the comparison test, your initial series has the same radius of convergence as 
$$
\sum_{n\geq 1} \frac{x^n}n =-\log (1-x)
$$ that is $R=1$.
A: $R=\lim {\dfrac{a_n}{a_{n+1}}}=\dfrac{na_n}{(n+1)a_{n+1}}\times \dfrac{n+1}{n}=1$ as $n\rightarrow \infty$
A: The root test will
also show that
$R=1$
since
$n^{1/n}
\to 1$
and
$1101^{1/n} \to 1$.
A: It can be proven that the power series and its derivative have the same radius of convergence. So consider the derivative of the above series $\displaystyle \sum_{n=1}^\infty na_nx^{n-1}, b_{n-1} = na_n$, and using root test:
$L = \displaystyle \lim_{n\to \infty} \sqrt[n-1]{\left|b_{n-1}x^{n-1}\right|}= |x|\cdot \displaystyle \lim_{n\to \infty} \sqrt[n-1]{na_n}= |x|$. And the series converges iff $|x| < 1$, thus $R = 1$.
