Do these converge?

Question

For the following choices for $a_k$, use the indicated test to show whether $\sum a_k$ converges or diverges

$\frac{1}{k^{1/k}k}$ (Comparison Test, limit form)

$\binom{2k}{k}^{-1}(4-10^{-23})^k$ (Ratio Test)

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For the first, I suspect it converges. Consider $\frac{1}{k^2}$

$\frac{\frac{1}{k^{1/k}k}}{\frac{1}{k^2}}=k^{\frac{1-k}{k}}$

Since $0 \leq\frac{1-k}{k}\leq 1$, one would suspect the limit of $\frac{\frac{1}{k^{1/k}k}}{\frac{1}{k^2}}=k^{\frac{1-k}{k}}$ to converge to at least zero as $k$ goes to $\infty$. Even if $L=0$, so long as $\sum\frac{1}{k^2}$ converges, then $\sum\frac{1}{k^{1/k}k}$ converges. Since the former summation converges, so to does the latter.

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For the second, I applied the Ration Test and did algebra to simplify. I got

$\frac{1+\frac{1}{k}}{2k+1}(4-10^{-23})^k$

Not sure what to do from here.

Answer 1

$$a_k=\binom{2k}k^{-1}\left(4-10^{-23}\right)^k=\frac{(k!)^2}{(2k)!}\left(4-\frac1{10^{23}}\right)^k\implies$$

$$\frac{a_{k+1}}{a_k}=\frac{((k+1)!)^2}{(2k+2)!}\left(4-\frac1{10^{23}}\right)^{k+1}\cdot\frac{(2k)!}{(k!)^2}\left(4-\frac1{10^{23}}\right)^{-k}=$$

$$\frac{(k+1)^2}{(2k+1)(2k+2)}\left(4-\frac1{10^{23}}\right)\xrightarrow[k\to\infty]{}\frac{4-\frac1{10^{23}}}{4}<1$$

and thus the series converges.

Answer 2

for $2$, $u_k<4^k$

$\frac{s_{k+1}}{s_k}=4>1$ Hence $s_k=4^k$ is convergent and $s_k $ dominates $u_k$