Radius of convergence of $\sum_{k=1}^{\infty} {{k^{\sqrt k}}x^{k}}$

Question

I'm currently computing sums to find the interval/radius of convergence.

But recently I'm stuck with this sum: $\sum_{k=1}^{\infty} {{k^{\sqrt k}}x^{k}}$.

I tried applying the ratio test $\lim_{k \to \infty}|\frac{a_{k+1}}{a_k}|$, thus: $\lim_{k \to \infty}|\frac{{(k+1)^{\sqrt {k+1}}}x^{k+1}}{{k^{\sqrt k}}x^{k}}|$.

Now we can write this as: $\lim_{k \to \infty}|\frac{{(k+1)^{\sqrt {k+1}}}x}{{k^{\sqrt k}}}|$.

I guess you could now write this as: $|x|\cdot\lim_{k \to \infty}|\frac{{(k+1)^{\sqrt {k+1}}}}{{k^{\sqrt k}}}|$.

Here I'm stuck since I'm not sure how to further simplify or solve the limit. I would be very glad if someone could show me how to solve the last part.

Answer 1

From the root test we have

$$\lim_{k\to \infty}\sqrt[k]{\left|k^{\sqrt k}x^k\right|}=|x|\lim_{k\to\infty}k^{1/\sqrt{k}}=|x|$$

Can you finish?

Answer 2

Following your approach by ratio test we have that

$$\frac{{(k+1)^{\sqrt {k+1}}}}{{k^{\sqrt k}}}=\frac{{(k+1)^{\sqrt {k+1}}}}{{k^{\sqrt {k+1}}}}\frac{{k^{\sqrt {k+1}}}}{{k^{\sqrt k}}}=\left(1+\frac1k\right)^{\sqrt {k+1}}k^{(\sqrt {k+1}-\sqrt k)}\to 1$$

indeed

$$\left(1+\frac1k\right)^{\sqrt {k+1}}=\left[\left(1+\frac1k\right)^k\right]^{\frac{\sqrt {k+1}}{k}}\to e^0=1$$

and

$$k^{(\sqrt {k+1}-\sqrt k)}=e^{(\sqrt {k+1}-\sqrt k)\log k}=e^{\frac{\log k}{\sqrt {k+1}+\sqrt k}}\to e^0=1$$

Answer 3

Similar to gimusi's answer, consider $$a_k=\frac{{(k+1)^{\sqrt {k+1}}}}{{k^{\sqrt k}}}\implies \log(a_k)={\sqrt {k+1}}\log(k+1)-\sqrt k \log(k)$$ that we can write as $$\log(a_k)=\sqrt k \sqrt{1+\frac{1}{k}}\left(\log(k)+\log \left(1+\frac{1}{k}\right) \right)-\sqrt k \log(k)$$ Now, using Taylor expansions and simplifying, we should end with

$$\log(a_k) =\frac{1+\frac 12 \log(k)}{\sqrt k}+O\left(\frac 1k\right)$$ $$a_k=e^{\log(a_k)}=1+\frac{1+\frac 12 \log(k)}{\sqrt k}+O\left(\frac 1k\right)$$