What is the limit of (a^n)/n for a>1?

[Delta2]

TL;DR

The limit of the sequence $a^n/n$ for $a>1$ is $+\infty$ but how to prove it without reverting to limit of functions and using L'Hopital rule.

We have the limit of the sequence $\frac{a^n}{n}$ where $a>1$. I know it is $+\infty$ and i can prove it by switching to the function $\frac{a^x}{x}$ and using L'Hopital.

But how do i prove it using more basic calculus, without the knowledge of functions and derivatives and L'Hopital.

I can prove that the sequence is increasing (after a certain $n_0$) but have trouble proving that the sequence is not bounded. More specifically
$$\frac{a^n}{n}>M\Rightarrow n\ln a-\ln n>\ln M$$ and where do i go from here?

 

[Infrared]

Let $b_n=a^n/n$. The ratio of successive terms $b_{n+1}/b_n=a\left(\frac{n}{n+1}\right)$ tends to $a>1$ as $n\to\infty$, so the sequence diverges to infinity (and by the "ratio test", this tells us the even stronger statement that $\sum_{n=1}^\infty b_n^{-1}$ converges.)

 

[Delta2]

Which theorem is that ?(if the ratio of successive terms has limit >1 then the sequence diverges) . I know the ratio test for series, but haven't heard of a ratio test for sequences.

 

[member 587159]

Which theorem is that ?(if the ratio of successive terms has limit >1 then the sequence diverges) . I know the ratio test for series, but haven't heard of a ratio test for sequences.

He uses the theorem you mention, together with the fact that if $\sum_n x_n$ converges, then $x_n \to 0$.

 

[Delta2]

He uses the theorem you mention, together with the fact that is $\sum_n x_n$ converges, then $x_n \to 0$.

Ok I see now thanks, due to ratio test $\sum \frac{1}{b_n}$ converges hence $\frac{1}{b_n}\to 0$ hence $b_n\to +\infty$ since $b_n$ is strictly positive.

 

[Infrared]

Yes, like @Math_QED said, it follows from the test for series. I don't know if it has a separate name for sequences, but it's easy enough to prove: Let $c_n$ be a sequence of positive terms, and suppose $c_{n+1}/c_n$ has limit(inf) $T>1.$ Pick $L\in (1,T)$, and let $N$ be so that $c_{n+1}/c_n>L$ for $n\geq N$. Then, for any $m>N$, we have $c_m=\left(\frac{c_m}{c_{m-1}}\right)\ldots\left(\frac{c_{N+1}}{c_N}\right)c_N> L^{m-N}c_N,$ and this tends to infinity. And this comparison is how you prove the test for series anyway, since geometric series diverge/converge according to their common ratio.

 

[mathwonk]

What about this? If a > 1, then write a = 1+h with h > 0.
Then by the binomial theorem, a^n > 1 + nh + n(n-1)/2h^2.
Hence a^n/n > 1/n + h + (n-1)/2. h^2 > (n-1)/2. h^2, which --> infinity as n does.

 

[WWGD]

This is another approach. Since a>1 , similar to Wonk, write it as 1+h; h>0. Then use logs or a rule of thumb like rule of 70 to find tge least power k with $(1+h)^k=2$ . Then you can see how for fixed values $n_i*k$ of $n$ , the ratio becomes:

$\frac{a^n}{n}=\frac{a^{n_i*k}}{n_ik}=\frac{2^{n_i}}{n_i*k}$ I think from here, comparing a doubling process with a linear growth process, with a bit of work you can show the expression goes to infinity.