Radius of Convergence(1)

Consider the function $\displaystyle\frac{x^k}{k^2}$ . Since exponentials grow faster than polynomials, this function diverges for $|x|>1$ as $k\to\infty$ .

Now consider $\displaystyle\left(\frac{x^k}{k^2}\right)^k$ . Since $|x^k|>|x|$ for $|x|>1,k>1$ , this function also diverges for $|x|>1$ as $k\to\infty$ . Now, letting $k=\sqrt n$ , we see that $\left(\frac{x^{\sqrt n}}{({\sqrt n})^2}\right)^{\sqrt n}=\frac{x^n}{n^{\sqrt n}}$ must also diverge, and hence the sum must diverge for $|x|>1$ . So, we know that the radius of convergence must be at least 1.

To prove that the sum converges for $|x|\leq1$ , we look at $x=1$ : $\begin{aligned}\sum_{n=0}^{\infty}\frac{1^n}{n^{\sqrt n}}&=1+1+\sum_{n=2}^{\infty}\frac1{n^{\sqrt n}}\\ &<1+1+\sum_{n=2}^{\infty}\frac1{n^{\sqrt2}}\end{aligned}$ By the p-series test, this sum converges, and hence the required sum converges for $x=1$ .

Then, if $|x|\leq1$ we have $|x^n|\leq1$ for any $n\geq0$ , and hence $\sum_{n=0}^{\infty}\left|\frac{x^n}{n^{\sqrt n}}\right|\leq\sum_{n=0}^{\infty}\frac1{n^{\sqrt n}}$ This proves that the sum is absolutely convergent, and therefore convergent, for all $|x|\leq1$ , and divergent for all $|x|\geq1$ , and thus the radius of convergence is 1.

Consider the series $\sum_{n=1}^{\infty}\frac{x^n}{n^{\sqrt{n}}}$ whose coefficients are $a_n= n^{-\sqrt{n}}$ ,

then let $\alpha = \lim \sup |a_n| ^{\frac{1}{n}}= \lim |n^{-\sqrt{n}}|^{\frac{1}{n}}= \lim \sup n^{\frac{-1}{\sqrt{n}}}$ , (sup: supreme )

since the power is always negative, all the terms must be less than or equal to 1; therefore, $\alpha \le 1$ .

Now, take the sub sequence of terms for $n = 2^k \implies n^{-1}{\sqrt{n}} = 2^{\frac{-k}{ 2^ { \frac{k}{2}}}}$ , as $k$ goes to infinity, $\frac{k}{2^{\frac{k}{2}}}$ goes to $0$ , implies $2^{\large\frac{-k}{2^{\frac{k}{2}}}}$ goes to $1$ . Since the limit of sub sequence is equal to $1$ , and we have proved earlier that the limit of the sequence is less than or equal to $1$ ; therefore, the limit of the series is $1$ , thus the radius of convergence is $1$ .