# Find the radius of convergence for the series $\sum_{n=1}^\infty c_nx^n$

Find the radius of convergence for the series
$\sum_{n=1}^\infty c_nx^n$,
where
$c_n= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}$

I know how to find the radius of convergence if $c_n$ was a specific function, but here it's not. I need to find the limit as $x \to \infty$ of $|\frac{c_{n+1}x^{n+1}}{c_nx^n}|$.
What am I supposed to do with this information $c_n= \frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}$ ?

• Are you sure $c_n$ is not $\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\ldots+\frac{1}{\sqrt{n^2+n}}$ ? I think you put parenthesis in the wrong places. Commented Nov 14, 2015 at 21:57
• The way you define $c_n$ isn't clear: is $c_n$ fixed in n, or is it dependent on another variable in the sum? Your general form says $\dfrac{1}{\sqrt{n^2 + n}}$, but n was defined generally before... Commented Nov 14, 2015 at 21:59
• You just need a couple of estimates for the $c_n$ to find the radius of convergence. What is the radius of convergence for $\sum n^{\alpha} x^n$? Commented Nov 14, 2015 at 22:00

$$\sum_{k=1}^{n}\frac{1}{\sqrt{n^2+k}}=\sqrt{n}\cdot\frac{1}{n}\sum_{k=1}^{n}\frac{1}{\sqrt{n+\frac{k}{n}}}\approx \sqrt{n}\int_{0}^{1}\frac{dx}{\sqrt{n+x}}=\frac{2\sqrt{n}}{\sqrt{n}+\sqrt{n+1}}\approx 1.$$

So we have that our coefficients approach one quite fast. That gives that the radius of convergence of the associated analytic function is one, too.

• Does this fully answer the question? This defines the limit of $c_n$ but not the original limit. Commented Nov 14, 2015 at 22:04
• @theREALyumdub: obviously it does. If the coefficient approach one quite fast the radius of convergence is one, too. Commented Nov 14, 2015 at 22:05
• I understand the connection, but also the question is unclear. Include that in your answer for clarity. Commented Nov 14, 2015 at 22:05
• @theREALyumdub: all right, done. Commented Nov 14, 2015 at 22:06
• I don't fully understand what method you used to find that $c_n$ approachs 1, is $c_1=\frac{1}{\sqrt{1^2+1}}$ and $c_2=c_1+\frac{1}{\sqrt{2^2+2}}$? I got $c_1=.707$ and $c_2=1.115$? Commented Nov 14, 2015 at 22:17

Note

$$\tag 1 \frac{n}{\sqrt {n^2 + n}} \le c_n \le \frac{n}{\sqrt {n^2 + 1}}$$

for all $n.$ Since the right and left sides of $(1)$ $\to 1,$ $c_n \to 1,$ which implies $c_n^{1/n} \to 1.$ By the root test, the radius of convergence is $1.$

• The Hadamard Radius formula : $\sum_na_nx^n$ converges for $|x|<r$ and diverges for $|x|>r$ where $r=1/(\lim \sup |a_n|^{1/n}).$ Commented Nov 15, 2015 at 0:01