Question

Determine the convergence or divergence of the sequence with the given nth term. If the sequence converges, find its limit. (If the quantity diverges, enter DIVERGES.) an = 1/sqrt(n)

Answer 1

This sequence converges to 0.

Proof: Recall that

$\displaystyle\lim_{n\to\infty}\frac1{\sqrt n}=0$

is to say that for any given $\varepsilon>0$, there is some $N$ for which $\left|\frac1{\sqrt n}-0\right|=\frac1{\sqrt n}$ for all $n>N$.

Let $N=\left\lceil\frac1{\varepsilon^2}\right\rceil$. Then

$n>\left\lceil\dfrac1{\varepsilon^2}\right\rceil\ge\dfrac1{\varepsilon^2}$

$\implies\dfrac1n$

$\implies\dfrac1{\sqrt n}$

as required.