Is this a correct way to show that $sum_{k=2}^{infty} frac {1}{sqrt k+k(-1)^k}$ converges?

Question

Show that $sum_{k=2}^{infty} frac {1}{sqrt k+k(-1)^k}$ converges or diverges. My attempt is ot first rewrite and see if the sequence is convergent according to the Leibniz test:
$frac {1}{sqrt k+k(-1)^k}= frac{(-1)^k}{sqrt k} frac{1}{sqrt k + (-1)^k} lefrac{(-1)^k}{sqrt k} frac{1}{sqrt k -1}$
Now see if it holds that:

$a_{k+1} le a_k$
$lim_{k rightarrow infty} a_k = 0$

The first criterion holds because the denominator is an increasing functions and thus an increasing sequence:
$f(x) = x-sqrt x implies f'(x)=1-frac{1}{2sqrt x} ge 0  forall x>2$
The second criterion also holds because: $lim_{k rightarrow infty} frac{1}{sqrt k} frac{1}{sqrt k-1} rightarrow 0 * 0=0 $. So according to the Leibniz test the sum converges

Mark Viola · Answer

It is true that $a_k=frac1{sqrt{k}+(-1)^kk}$ is less than $b_k=frac{(-1)^k}{sqrt{k}(sqrt{k}-1)}$.  But each of these sequences is alternating in sign.  And even if $sum_{k=2}^infty b_k$ converges, the dominance of $b_k$ over $a_k$ does not guarantee the convergence of $sum_{k=2}^infty a_k$.
However, we can proceed as follows.  Note that we can write
$$begin{align}
frac1{sqrt{k}+(-1)^kk}&=frac{sqrt{k}-(-1)^kk}{k(1-k)}\
&=frac{1}{sqrt{k}(1-k)}-frac{(-1)^k}{1-k}
end{align}$$
Hence, we can write
$$begin{align}
sum_{k=2}^K frac1{sqrt{k}+(-1)^kk}&=sum_{k=2}^K frac{1}{sqrt{k}(1-k)}-sum_{k=2}^Kfrac{(-1)^k}{1-k}tag1
end{align}$$
The first sum on the right-hand side of $(1)$ converges as $Kto infty$ by comparison to $sum_{k=1}^infty frac1{k^{3/2}}$ and the second sum converges as guaranteed by Leibniz's Test.

Is this a correct way to show that $sum_{k=2}^{infty} frac {1}{sqrt k+k(-1)^k}$ converges?

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