# Thread: Having some trouble with a limit

1. ## Having some trouble with a limit

Dear all,

I'm having some trouble with the following limit:
n(x^n) as n tends to infinity, knowing that |x| < 1
I need it to calculate a series.

Please note that I am not allowed to use exp and ln as we haven't covered that in the syllabus yet. However, I'd be glad to know of a solution using these functions too.

Yack

PS: this is my first post here, sorry if I left this message in the wrong place.

2. Originally Posted by Yack
Dear all,

I'm having some trouble with the following limit:
n(x^n) as n tends to infinity, knowing that |x| < 1
I need it to calculate a series.

Please note that I am not allowed to use exp and ln as we haven't covered that in the syllabus yet. However, I'd be glad to know of a solution using these functions too.

Yack

PS: this is my first post here, sorry if I left this message in the wrong place.
what do you want exactly? to calculate a limit, or to tell if a series converges? it looks to me like the root test is your best bet for the latter. the ratio test can work too

3. Originally Posted by Jhevon
what do you want exactly? to calculate a limit, or to tell if a series converges? it looks to me like the root test is your best bet for the latter. the ratio test can work too

I'm trying to calculate the limit of the sequence n*x^n.

4. Originally Posted by Yack
I'm trying to calculate the limit of the sequence n*x^n.
Look at the root, $|nx^n|^{1/n} = n^{1/n} |x|$ now $n^{1/n} \to 1$ thus $n^{1/n} |x| \to |x|$. Thus, if $|x|<1$ then the sequence limit is $0$. If $|x|>1$ then the sequence limit diverges. And if $|x|=1$ then we have two cases to check $x=\pm 1$ in either case the limit does not exist.

5. Originally Posted by ThePerfectHacker
Look at the root, $|nx^n|^{1/n} = n^{1/n} |x|$ now $n^{1/n} \to 1$ thus $n^{1/n} |x| \to |x|$. Thus, if $|x|<1$ then the sequence limit is $0$. If $|x|>1$ then the sequence limit diverges. And if $|x|=1$ then we have two cases to check $x=\pm 1$ in either case the limit does not exist.
Thanks a lot. This makes it much more obvious. However I'm having trouble to find a formal proof (using Cauchy sequences, the squeeze theorem, or the definition of limits).

Yack

6. Originally Posted by Yack
Thanks a lot. This makes it much more obvious. However I'm having trouble to find a formal proof (using Cauchy sequences, the squeeze theorem, or the definition of limits).

Yack
Theorem: Let $a_n$ be a sequence such that $|a_n|^{1/n}$ is a convergent sequence with limit $a$. If $a<1$ then the sequence $a_n \to 0$ and if $a>1$ then $a_n$ is divergent.

Proof: Say $0\leq a<1$ (it cannot be that $a<0$ because this is a sequence of non-negative terms) and $\lim ~ |a_n|^{1/n} = a$ then $\left| |a_n|^{1/n} - a \right| < \epsilon$ for $n\geq N$. This means, $a - \epsilon < |a_n|^{1/n} < a+\epsilon$ for $n\geq N$, choose $\epsilon$ so small that $a+\epsilon < 1$ (which is possible since $a<1$). Then it means $0\leq |a_n|^{1/n} \leq (a+\epsilon)\implies 0\leq |a_n| \leq (a+\epsilon)^n \mbox{ for }n\geq N$ by the Squeeze theorem $\lim ~ |a_n| = 0$ because $(a+\epsilon)^n \to 0$. Thus, $\lim ~ a_n = 0$. Q.E.D.

The case with $a>1$, is similar, i.e. pick an $\epsilon$ so small that $a-\epsilon > 1$ and then $(a-\epsilon)^n \leq |a_n|$.