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.

  • $\begingroup$ Are these homework problems? They do remind me of the questions I saw when I was a freshman. $\endgroup$ – Asaf Karagila Apr 25 '11 at 12:02
  • $\begingroup$ They were my homewrok couple of years ago, yes. $\endgroup$ – user6163 Apr 25 '11 at 12:26

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 $0\leq a_n\leq \left(\frac{1+l}2\right)^n$.

  • $\begingroup$ The magical epsilon. thanks! :-) $\endgroup$ – user6163 Apr 25 '11 at 11:34

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 $a_n \lt b^n \to 0$.


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