This sequence converges to 0.
Proof: Recall that
[tex]\displaystyle\lim_{n\to\infty}\frac1{\sqrt n}=0[/tex]
is to say that for any given [tex]\varepsilon>0[/tex], there is some [tex]N[/tex] for which [tex]\left|\frac1{\sqrt n}-0\right|=\frac1{\sqrt n}<\varepsilon[/tex] for all [tex]n>N[/tex].
Let [tex]N=\left\lceil\frac1{\varepsilon^2}\right\rceil[/tex]. Then
[tex]n>\left\lceil\dfrac1{\varepsilon^2}\right\rceil\ge\dfrac1{\varepsilon^2}[/tex]
[tex]\implies\dfrac1n<\varepsilon^2[/tex]
[tex]\implies\dfrac1{\sqrt n}<\varepsilon[/tex]
as required.
