# $\lim_{n\to \infty }\sqrt{a_{n}}< 1$, $a_{_{n}}\geq 0$ for every $n \in \mathbb{N}$- prove that $a_{n}$ is convergent

It is very similar to this question that I posted not a while ago,

But I’m still having a hard time to transalte or use the solution that was given.

Now ,the sequence $a_{_{n}}$ applies these condition:

$a_{_{n}}\geq 0$ for every $n \in \mathbb{N}$

$\lim_{n\to \infty }\sqrt[n]{a_{n}}< 1$

again,I need to prove that $a_{n}$ is convergent,
and it’s limit is 0.

As was suggested I tried to follow the limit defenition.

#### Solutions Collecting From Web of "$\lim_{n\to \infty }\sqrt{a_{n}}< 1$, $a_{_{n}}\geq 0$ for every $n \in \mathbb{N}$- prove that $a_{n}$ is convergent"

Put $l:=\lim_{n\to+\infty}\sqrt[n]{a_n}$. We apply the definition of the limit for $\varepsilon =\frac{1-l}2>0$. We can find $N\in\mathbb N$ such that if $n\geq N$ then $\sqrt[n]{a_n}\leq l+\frac{1-l}2 = \frac{1+l}2$. We get for $n\geq N$ that $0\leq a_n\leq \left(\frac{1+l}2\right)^n$.

That’s the root test for the convergence of the series $\sum a_n$. If the series converges, then its terms $a_n$ converge to zero.

Alternatively, you can do it directly: if $\lim_{n\to \infty }\sqrt[n]{a_{n}} \lt 1$ then there is $b$ with $0 < b < 1$ such that $\sqrt[n]{a_{n}} \lt b$ for $n$ large enough. Then $a_n \lt b^n \to 0$.