I have the following sequence $$S_n=1+\sum_{m=2}^n \prod_{k=m}^n a_k$$ where $\{a_k\}$ is a real positive sequence with the property that $$\lim_{n\to \infty}\left(\prod_{k=1}^n a_k\right)^{1/n}<1$$ Had the product in the definition of the sequence started from $k=1$ instead of $k=m$, the $S_n$'s would have been partial sums of a series and by Cauchy's root test I could say that the sequence $S_n$ converges and hence $S_n$ would have been bounded. But this type of definition of the sequence has made it difficult to check whether it is bounded or not. Can anybody give me some hint regarding how to proceed in this case? Thanks in advance.

Edit: I have had the following idea regarding this problem. Is this method of proof correct? Please give me some suggestions regarding this.

We can see that $$S_n=1+a_n+a_na_{n-1}+\cdots+a_{n}a_{n-1}\cdots a_2$$ Let us define a sequence $\{b^{n}_k\}_{k\ge 1}$ such that $b^{n}_k=a_{n-k+1},\ k=1,2,\cdots,\ n-1$ and $b^n_{k}=a_{k+1},\ k\ge n$. Then, $$S_n=1+b^n_1+b^n_1b^n_2+\cdots+b^n_{1}b^{n}_2\cdots b^n_{n-1}$$ which is a partial sum of the series $1+\sum_{k\ge 1}\prod_{j=1}^k b^n_j$ which is convergent by Cauchy's root test. Hence $S_n$ is bounded.

  • $\begingroup$ @Elaqqad Ok, I understand the logic. Thanks for that. Also can you check if the proof I have supplied is correct? $\endgroup$ – Samrat Mukhopadhyay Aug 11 '15 at 10:25
  • $\begingroup$ @Elaqqad, I think it follows from the condition that the $a_n$'s have. $\endgroup$ – Samrat Mukhopadhyay Aug 11 '15 at 10:35
  • $\begingroup$ the sum $\sum_{k\ge 1}\prod_{j=1}^k b^n_j$ is not of the form $\sum_{k\ge 1}f(k)$ but it's of the form $\sum_{k\ge 1} f_n(k)$ which is convergent using the condition for every $n$ to some limit $l_n$ but it does not coincide with $S_n$, to summurize : Your $S_n$ is not the partial sum of the given serie $\endgroup$ – Elaqqad Aug 11 '15 at 10:43
  • $\begingroup$ @Elaqqad, let me explain this carefully. What I am trying to say is that for each $n,\ $ $S_n$ can be written as a partial sum of a series that converges by Cauchy's root test. Hence, for each $n,$ $S_n$ is bounded, though the different $S_n$'s are partial sums to different series all of which converges to some limits $l_n$. So, $l_n$ is a sequence of positive real numbers and $S_n<l_n$. Does this sound okay? $\endgroup$ – Samrat Mukhopadhyay Aug 11 '15 at 10:51
  • $\begingroup$ You did not get it yet!, let's write $T_n=\sum_{k\ge 1} f_n(k)=\sum_{k\ge 1}\prod_{j=1}^k b^n_j $(this serie is convergent for every $n$ to some limit $l_n$ because $f_n(k)^{1/k}$ tends to some integer less than $1$, try to prove this , it's not easy!!!). This does not have to do anything with $S_n$!!, the only relation is the fact that $S_n=T_n(n)$, so every $S_n$ is only a term in $T_n$, and $S_{n+1}$ is a term in $T_{n+1}$ this is the meaning of what you did, from this you can possibly conclude that $S_n\leq l_n$ but the question would be is $l_n$ bounded ... $\endgroup$ – Elaqqad Aug 11 '15 at 11:04

Let $a_n = \exp( -1 + (-1)^n \sqrt{n} )$. Then we know that $(a_1 \cdots a_n)^{1/n} \to e^{-1}$ as $n \to \infty$. On the other hand, we also know that

$$ S_{2n} \geq a_{2n} = e^{\sqrt{2n}-1} \longrightarrow +\infty \quad \text{as} \quad n \to \infty. $$

So $(S_n)$ cannot converge in this case.

More generally, notice that $S_n = 1 + a_n S_{n-1}$. Assuming that $(S_n)$ converges to $\sigma$, we should have that $(a_n)$ converges:

$$ a_n = \frac{S_n - 1}{S_{n-1}} \xrightarrow[n\to\infty]{} \frac{\sigma - 1}{\sigma} <1 $$

The converse is very straightforward, so we find that

$$ (S_n) \text{ converges} \quad \Longleftrightarrow \quad (a_n) \text{ converges to a limit in } [0, 1). $$

  • $\begingroup$ Oh, ok, right, that example you told me earlier. $\endgroup$ – Samrat Mukhopadhyay Aug 11 '15 at 11:04
  • $\begingroup$ @SamratMukhopadhyay The same counterexample as in the answer for your first question works also you did not just check it !! ($a_{2n}=n/3$ and $a_{2n+1}=2/n$ $\endgroup$ – Elaqqad Aug 11 '15 at 11:15
  • $\begingroup$ Now I know @Elaqqad. $\endgroup$ – Samrat Mukhopadhyay Aug 11 '15 at 11:47
  • $\begingroup$ Sangchul, I think $S_n$ remains bounded, though may not converge if $a_n$ does not converge and we have $\lim\sup a_n<1$, right? $\endgroup$ – Samrat Mukhopadhyay Aug 17 '15 at 14:27
  • $\begingroup$ @SamratMukhopadhyay, Yes, boundedness of $(S_n)$ holds in a more general context (such as in the case you suggested) and the equivalent condition for $(a_n)$ seems much less obvious... I have thought this problem for a while, but then I gave up. $\endgroup$ – Sangchul Lee Aug 17 '15 at 15:24

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