Let $(a_n)_{n\in\mathbb{N}}$ be a sequence on $\mathbb{R}$, and let $p$ be a polynomial with positive leading coefficient. Let me state a definition first:

Divergence of a sequence to $\pm\infty$

A sequence $s_n$ diverges to $+\infty$ if and only if for any $M>0$ we can find an $N\in\mathbb{R}$ such that for any $n>N$, we have $a_n>M$. A sequence $s_n$ diverges to $-\infty$ if and only if for any $M<0$ we can find an $N\in\mathbb{R}$ such that for any $n>N$, we have $a_n<M$.

So by this definition, $a_n=(-1)^n$ is not a diverging sequence.

Let me now propose the following conjecture:

The sequence $$\frac{a_{n+1}-p(n)}{a_1+a_2+\cdots+a_n+p(n)}$$ diverges to $+\infty$ if and only if the sequence $$\frac{a_{n+1}}{a_1+a_2+\cdots+a_n}$$ diverges to $+\infty$.

My thoughts.

The one side is obvious. We see that, for large enough $n$, the polynomial $p$ takes only positive values, so that


so if the left-hand side diverges, then so does the left side.

The other way however is a little more challenging. I feel like it is true, since $a_n$ must be diverging really fast for the above two expressions to diverge. For example, even an exponential sequence $2^n$ is not diverging fast enough, since:

$$\lim_{n\to\infty} \frac{2^{n+1}}{2^1+2^2+\cdots+2^n}=\lim_{n\to\infty} \frac{2^{n+1}}{2^{n+1}-2}=1$$

but $2^{n^2}$ is enough, since \begin{align} \lim_{n\to\infty} \frac{2^{(n+1)^2}}{2^{1^2}+2^{2^2}+\cdots+2^{n^2}}&\geq \lim_{n\to\infty} \frac{2^{(n+1)^2}}{1+2^0+2^1+2^2+2^3+\cdots+2^{n^2}}\\ &=\lim_{n\to\infty} \frac{2^{n^2+2n+1}}{2^{n^2+1}}\\ &=\lim_{n\to\infty} 2^{2n}\\ &=+\infty \end{align} So in the intuitive sense: "$p(n)$ is negligible in comparison to $a_{n+1}$ and $a_1+\cdots+a_n$". How can I prove this rigorously?

  • $\begingroup$ A series is divergent if it does not converge. $(-1)^n$ does not converge, so it is divergent. As you say, it doesn't diverge to $\infty$ or $-\infty$, but it is still divergent. $\endgroup$ – Empy2 Mar 10 '16 at 10:35
  • 1
    $\begingroup$ @Michael, I know that, but that's not the divergence I am talking about here. I stated the definition for "divergence to $\pm\infty$" not because I thought that's the definition of divergence, but to avoid any misconceptions. I edited the "divergence of a sequence" header to "divergence of a sequence to $\pm\infty$". I hope this clears things up. $\endgroup$ – vrugtehagel Mar 10 '16 at 10:41
  • $\begingroup$ @Michael There are some countries (e.g. Italy, from where I come) where a divergent sequence has another definition: i.e. it is a sequence diverging to $\infty$ (this is very standard in Italy, I have read it from many sources). Sequences having no limit here are called "indeterminate sequences". $\endgroup$ – Crostul Mar 10 '16 at 10:51
  • $\begingroup$ Yes. $\frac{a_{n+1}}{a_1+a_2+\cdots+a_n}$ diverging means that $a_n$ grows faster than any exponential $a^n$ because for exponential the quantity is bounded (it's around $\alpha-1$). @cros I think reserving divergent for sequences going off to infinity is justified (and calling others non-convergent) just as saying that a moment is defined even if its infinite. $\endgroup$ – A.S. Mar 10 '16 at 10:51

To prove the other direction, suppose that $a_{n+1}/(a_1+...+a_n)$ diverges to $+\infty$.

Choose some $N$ large enough so that $n \geq N$ implies that $a_{n+1}/(a_1+...+a_n) \geq 2$. Then we see that $a_{N+1} \geq 2(a_1+...+a_N)$. Thus $a_{N+2} \geq 2(a_1+...+a_{N+1}) \geq 6(a_1+...+a_N)$. Continuing, you obtain that $a_{N+3} \geq 18(a_1+...+a_N)$. Proceeding inductively you can prove that $$a_{N+k} \geq 2 \cdot 3^{k-1}(a_1+...+a_N)$$

Thus if $p$ is any polynomial, then for $n \geq N$ we know that $\big|\frac{p(n)}{a_n}\big| \leq \frac{1}{2 \cdot 3^{-N-1} |a_1+...+a_N|}\big|\frac{p(n)}{3^n}\big|$ which tends to zero as $n$ tends to $\infty$. Thus given $M>0$ we choose some $K$ such that $n \geq K$ implies that $|p(n)| \leq |a_n|/2$ and $a_{n+1}/(a_1+...+a_n) \geq 4M$. Then $n \geq K$ implies that $$a_{n+1}-p(n) \geq a_{n+1}/2 \geq 2M(a_1+...+a_n) \geq M(a_1+...+a_n+p(n))$$ and since $M$ was arbitrary, the result follows.

  • $\begingroup$ Very nice! A view into the problem of divergent series which I've missed when I read K.Knopp on divergent series. (+1) $\endgroup$ – Gottfried Helms Mar 16 '16 at 17:14

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