$ 1+(1/\sqrt{3})-(a/\sqrt{2})+(1/\sqrt{5})+(1/\sqrt{7})-(a/\sqrt{4})+(1/\sqrt{9})+(1/\sqrt{11})-(a/\sqrt{6})... $ Converges By comparison test? For what value of real constant $a$ does the following series converge?
$$
1+(1/\sqrt{3})-(a/\sqrt{2})+(1/\sqrt{5})+(1/\sqrt{7})-(a/\sqrt{4})+(1/\sqrt{9})+(1/\sqrt{11})-(a/\sqrt{6})...
$$
I do not have a clue on how to proceed. Suggest a possible route.
can we do it by comparison test somehow? Can we comapre it with   $c/\sqrt{(n)}$ for some $c$ and some large enough $n$?

Okay, I think I can guess at the pattern with the most recent addition
$$
\begin{align}
&\sum_{n=0}^\infty\left(\frac1{\sqrt{4n+1}}+\frac1{\sqrt{4n+3}}-\frac{a}{\sqrt{2n+2}}\right)\\
&=\sum_{n=0}^\infty\frac{\sqrt{(4n+3)(2n+2)}+\sqrt{(4n+1)(2n+2)}-a\sqrt{(4n+1)(4n+3)}}{\sqrt{(4n+1)(4n+3)(2n+2)}}\\
&=\sum_{n=0}^\infty\frac{\sqrt{(1+\frac3{4n})(1+\frac1n)}+\sqrt{(1+\frac1{4n})(1+\frac1n)}-a\sqrt2\sqrt{(1+\frac1{4n})(1+\frac3{4n})}}{2\sqrt{n}\sqrt{(1+\frac1{4n})(1+\frac3{4n})(1+\frac1n)}}\\
&=\sum_{n=0}^\infty\frac{\left[1+O(\frac1n)\right]+\left[1+O(\frac1n)\right]-a\sqrt2\left[1+O(\frac1n)\right]}{2\sqrt{n}\left[1+O(\frac1n)\right]}\\
&=\sum_{n=0}^\infty\frac{2-a\sqrt2+O(\frac1n)}{2\sqrt{n}}\tag{1}
\end{align}
$$
where $O(\cdot)$ is Landau big-O notation. Since convergence concerns the tail of the sequence, we can replace the term for $n=0$ by it value, $1+\frac1{\sqrt3}-\frac{a}{\sqrt2}$, and start using $O(\frac1n)$ for $n\ge1$.
The sum in $(1)$ will converge if and only if $a=\sqrt2$.
