Does $\lim \frac{a_n}{e^{\delta n}}=0$ for every positive $\delta$ imply that $\lim \frac{a_n}{\sum\limits_{k=1}^n a_k} =0$

Question

Let's say we have an increasing sequence $a_n$ such that $\underset{n\rightarrow\infty}{\lim} a_n=\infty$. Now it's fairly clear to me, though I haven't proven this yet, that: $$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=K>0 \Rightarrow \exists\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=1$$ In fact, I think $\delta=1/K$ but I could have that mixed up. And as a corrollary: $$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=0 \Rightarrow \forall\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=0$$

But here's my question: is the converse also true? Specifically is it the case that: $$ \forall\ \delta>0, \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=0 \Rightarrow \underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=0 $$

I think that this is true but when I try to show it, the problem comes up that I can show that $a_n$ is dominated by the same sequence that dominates $\sum\limits_{k=1}^n a_k$, but I can't show that $\sum\limits_{k=1}^n a_k$ is large relative to $a_n$. In my mind though, I know that if the RHS is not true and this limit is positive, then at some point the sequence is proportional to its own rate of change, and that means it should be asymptotically equivalent to an exponential function. But the details of the proof are killing me.

Any help would be greatly appreciated :-)

Answer 1

Here is a counterexample to your first conjecture, stating that for every unbounded increasing sequence $(a_n)$, $$\underset{n\rightarrow\infty}{\lim} \frac{a_n}{\sum\limits_{k=1}^n a_k}=K>0 \Rightarrow \exists\ \delta>0: \underset{n\rightarrow\infty}{\lim} \frac{a_n}{e^{\delta n}}=1$$ Take $a_n=n2^n$. Then $\{a_n\}$ is increasing and $a_n\to\infty$. Then $$ a_1+\cdots+a_n=(2-1)(a_1+\cdots+a_n)=2^2+2\cdot 2^3+\cdots+n2^{n+1}-2^1-2\cdot 2^2-\cdots-n2^{n}=n2^{n+1}-2-(2^2+2^3+\cdots+2^n)=n2^{n+1}-2-2^{n+1}+2^2=(n-1)2^{n+1}+2. $$ Also $$ \frac{a_n}{a_1+\cdots+a_n}=\frac{n2^n}{(n-1)2^{n+1}+2}\to \frac{1}{2} $$ But, there is no $\delta>0$, such that $$ \frac{a_n}{\mathrm{e}^{n\delta}}=\frac{n2^n}{\mathrm{e}^{n\delta}}= \exp(\log n+n(\log 2-\delta)) $$ converges to $1$, as $\log n+n(\log 2-\delta)\to \infty$ or $\log n+n(\log 2-\delta)\to -\infty$, if $\delta\ge\log 2$ or $\delta<\log 2$, respectively.

Answer 2

Not true.

Take $a_n = n * (\frac{e}{2})^{n}$. It's easy to verify that $\lim \frac{a_n}{e^n} = 0$

On the other hand, $\frac{a_n}{\sum a_n} = \frac{1}{\sum a_i/a_n}$, where i runs from 1 to n

Now consider the denominator:

$\sum \frac{a_i}{a_n} = \sum \frac{i}{n}*(\frac{e}{2})^{i-n} $ < $\sum (\frac{e}{2})^{i-n}$

Let $k = \frac{e}{2}$. Then, the RHS can be written as: $\frac{1}{k^n}*\sum k^i$ = $\frac{k-1/k^n}{k-1} < \frac{k}{k-1} = \frac{e}{e-2}$

Thus, $\frac{a_n}{\sum a_n} = \frac{1}{\sum a_i/a_n} > \frac{e-2}{e} = constant > 0 $.

Then it can't go to $0$.

Answer 3

Let $a_k=e^{\beta k}$ with $0<\beta<\delta$ so that your assumption holds. Then $$\sum_1^n a_k=\frac{e^{(n+1)\beta}-e^{\beta}}{e^{\beta}-1}$$ now note that $$\lim_{n \to \infty} \frac{e^{\beta n}}{\frac{e^{(n+1)\beta}-e^{\beta}}{e^{\beta}-1}}=1-e^{-\beta}$$

Answer 4

Not true.

Take the sequence $\{a_n\}$ of the form: $$ a_n=b_j,\quad \text{for}\,\,\,k_j\le n< k_{j+1}, $$ such that

a. The $b_j$ are positive and form an increasing sequence,

b. $b_j=a_1+\cdots+a_{k_j-1}$, and hence $\dfrac{a_{k_j}}{a_1+\cdots+a_{k_j}}=\dfrac{1}{2}$,

c. $k_j$ is chosen so that $\dfrac{a_{k_j-1}}{a_1+\cdots+a_{k_j-1}}<\dfrac{1}{2^j}$.

d. Such sequence is: $1,1,2,2,2,8,8,8,8,8,8,8,64,\cdots,64,2^{10},\cdots$, which satisfies $$ a_{2^n} \le 2^{n(n+1)/2}\quad \text{and}\quad a_m\le m^{\log m}, $$ and hence $$ 0<\frac{a_m}{\mathrm{e}^{\delta m}}\le\frac{m^{\log m}}{\mathrm{e}^{\delta m}}\to 0, \quad\text{for all $\delta>0$.} $$