Does $\sum^{∞}_{k=1} \frac{4k^k}{2^{k^2}}$ converge absolutely?

Question

I was wondering if you guys could judge my reasoning and let me know if am correct in finding if this series converges absolutely or conditionally.

$$\sum^{∞}_{k=1} \frac{4k^k}{2^{k^2}}$$

I decided to use the root test because of the k's in the exponent.

$$\lim_{n → ∞} {(\frac{|4n^n|}{|2^{n^2}|}})^{1/n} → \lim_{n → ∞} {\frac{|4^{1/n}n^{n/n}|}{|2^{n^2/n}|}} → \lim_{n → ∞} \frac{|n|}{|2^{n}|}$$

Since the $\lim_{n → ∞} \frac{|n|}{|2^{n}|} = 0$ by growth rates, $\sum^{∞}_{k=1} \frac{|4k^k|}{|2^{k^2}|}$ converges. Which means $\sum^{∞}_{k=1} \frac{4k^k}{2^{k^2}}$ converges absolutely.

Answer 1

Yes, it converges absolutely.

A few points:

• Instead of the arrow $\rightarrow$, it is more common to use the equal sign $=$.

• You implicitly use the following results: $$ \lim_{n\to\infty}\sqrt[n]{4}=1,\quad \lim_{n\to\infty}\frac{n}{2^n}=0\;, $$ which is okay if you know why they are true.

• You could omit the absolute value in various places since every term is non-negative.