I'm given two sequences:


as well as an initial condition $a_1=1$, $b_1=2$, and am told to find: $\displaystyle \lim_{n\to\infty}{a_n}$.

Given that I'm not even sure how to approach this problem, I tried anyway. I substituted $b_{n-1}$ for $b_n$ to begin the search for a pattern. This eventually reduced to:


Seeing no pattern, I did the same once more:


While this equation is atrocious, it actually reveals somewhat of a pattern. I can sort of see one emerging - though I'm unsure how I would actually express that. My goal here is generally to find a closed form for the $a_n$ equation, then take the limit of it.

How should I approach this problem? I'm totally lost as is. Any pointers would be very much appreciated!


While there is a way to prove that $\displaystyle\lim_{n\to\infty}{a_n}=5$ using $\displaystyle f(x)=\frac{1}{x-1}$, I'm still looking for a way to find the absolute form of the limit, $\displaystyle\frac{1+2a+ab}{b-a}$.

  • $\begingroup$ Let $a_1 = a$ and $b_1 = b$ with the iterations defined as in the question. Suppose $b>a$. It seems that $\lim_{n\rightarrow\infty} a_n = \frac{ab+2a+1}{b-a}$ and $\lim_{n\rightarrow\infty} b_n = \infty$. For $a=b$, both $a_n$ and $b_n$ appear to diverge. Where did you find this question by the way? $\endgroup$ – Lord Soth May 25 '13 at 1:56
  • $\begingroup$ @Lord The series does converge, though - I know that much. I'm not sure where your error is, though. How did you arrive at that fractional representation? Also, it's for a math project; my teacher gave us a pile of difficult problems. $\endgroup$ – Aza May 25 '13 at 2:01
  • $\begingroup$ Yes, it converges with your initial conditions $a_1=1$, $b_1=2$, which will give an answer of $5$ with the fractional representation (put $a=1$ and $b=2$ to that formula). I found that prospective formula numerically, just in case it may give an idea on how to approach the problem. If $a_1 = 1$ and $b_1=1$ though (for example), both $a_n$ and $b_n$ appear to diverge. $\endgroup$ – Lord Soth May 25 '13 at 2:14
  • $\begingroup$ To put in other words, if you gave us $a_1 = 3$ and $b_1 = 4$ as initial conditions, then the limit of $a_n$ would be $\frac{4\times 3+2\times 3 + 1}{4-3} = 19$, and $b_n$ would diverge. $\endgroup$ – Lord Soth May 25 '13 at 2:18
  • $\begingroup$ @Lord Hmm, thanks! Hopefully this gives me the lead I need to solve this problem. I'll think about this some more! $\endgroup$ – Aza May 25 '13 at 2:28

To proceed

Actually we can compute $\lim a_n$ explicitly, since $$ \frac{c_{n+1}}{c_n}=\frac{1}{c-\frac{1}{c_n}} \text{ converges to } 1 \quad \Leftrightarrow \quad c_n \text{ converges to } \frac{1}{c-1} $$ which is also equivalent to that $ a_n \text{ converges to } \frac{1}{c-1}-1 $, where $c=\frac{c_2}{c_3}+\frac{1}{c_2}$.

We know$\ c_2=a_2+1=\frac{(1+a_1)(1+b_1)}{b_1},c_3=a_3+1=\frac{(+a_2)(1+b_2)}{b_2}=\frac{(1+a_1)^2(1+b_1)^2}{b_1(1+b_1+a_1b_1)}$

Now we get $c=\frac{1+2b_1+a_1b_1}{(1+a_1)(1+b_1)}$ and $$ a_n\to \frac{1}{c-1}-1=\frac{1+2a_1+a_1b_1}{b_1-a_1} $$ Edit:

Thanks to Mihai Dicu, once we notice that $\frac{1}{1+a_{n+1}}-\frac{1}{1+b_{n+1}}=\frac{1}{1+a_{n}}-\frac{1}{1+b_{n}}=\frac{1}{1+a_{1}}-\frac{1}{1+b_{1}}$, it is rather easy to find the limit. From my previous answer, if $b_1>a_1>0$, we can actually show that both $a_n$ and $b_n$ are stictly increasing, and $b_n\to +\infty$. Therefore, $$ \frac{1}{1+a_n}=\frac{1}{1+b_n}+\frac{1}{1+a_1}-\frac{1}{1+b_1}>\frac{1}{1+a_1}-\frac{1}{1+b_1}>0 $$ which shows that ${a_n}$ is bounded, and thus its limits exists. $$ \lim_{n\to \infty}b_n=+\infty \Rightarrow \lim_{n\to \infty}\frac{1}{1+a_n}=\frac{1}{1+a_1}-\frac{1}{1+b_1}=\frac{b_1-a_1}{(1+a_1)(1+b_1)} $$ which is equivalent to $$ \lim_{n\to \infty}a_n=\frac{(1+a_1)(1+b_1)}{b_1-a_1}-1=\frac{1+2a_1+a_1b_1}{b_1-a_1} $$

  • $\begingroup$ @Knight who say Ni Now are you satisfied with this answer?(I mean the edit, it's a better way to compute the limit indeed, but still cannot find a genreal formula for $a_n$ or $b_n$) $\endgroup$ – Coiacy Jun 11 '13 at 19:25
  • $\begingroup$ (Sorry I didn't reply sooner; I wasn't pinged!) Yeah, this actually works well! I can see the solution now. Thank you so much! $\endgroup$ – Aza Jun 11 '13 at 22:38
  • $\begingroup$ @KnightswhosayNi Frankly speaking, I like your question very much and thanks for the acceptance. $\endgroup$ – Coiacy Jun 12 '13 at 3:42
  • $\begingroup$ No problem! I have enjoyed fiddling with these equations as well, and have found your answer fascinating. $\endgroup$ – Aza Jun 12 '13 at 3:49

It's obvious that $a_n$ and $b_n$ are in the same situation, so their limits highly depend on the initial values. Following are some points we can obtain from $a_1=1$ and $b_1=2$:

  • $a_n>0$ and $b_n>0\ ;a_{n+1}-a_{n}=\frac{1+a_n}{b_n}>0$ and similarly $b_{n+1}-b_n>0$. Therefore, $\{a_n\}$ and $\{b_n\}$ are strictly increasing sequences;

  • $b_{n+1}-a_{n+1}=\frac{(b_n-a_n)+(b_n^2-a_n^2)+a_n\cdot b_n(b_n-a_n)}{a_n\cdot b_n}$, and thus by induction $b_n>a_n$ for every $n$;

  • $b_{n+1}-b_{n}=\frac{1+b_n}{a_n}>\frac{b_n}{a_n}>1$, which implies $b_n$ increases to $+\infty$;
  • $\frac{a_{n+1}}{a_n}=1+\frac{1}{b_n}+\frac{1}{a_n\cdot b_n}$ converges to $1$ as $n\to \infty$.

Now we prove $\lim a_n$ exists and find its closed form. To show the existence, it suffices to show $\{a_n\}$ is bounded. First assume $a_n$ increases to infinity, and we will derive a contradiction with the last point listed above. From the fact $$b_n(a_{n+1}+1)=(1+a_n)(1+b_n)=a_n(b_{n+1}+1)$$ Denote $c_n:=a_n+1$ and $d_n:=b_n+1$, then we obtain $$ \begin{cases} \frac{c_n-1}{c_n}=\frac{d_n}{d_{n+1}}\\ \frac{d_n-1}{d_n}=\frac{c_n}{c_{n+1}} \end{cases} \Rightarrow \begin{cases} \frac{d_{n+1}}{d_n}=\frac{c_n}{c_n-1}\\ \frac{c_{n+1}}{c_n}=\frac{d_n}{d_n-1} \end{cases} $$ For $n\ge 2$, $d_n=d_1\cdot \frac{d_2}{d_1}\cdots \frac{d_n}{d_{n-1}}=d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}$, which implies $$ d_1(1-\frac{c_n}{c_{n+1}})=(1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n-1}}) $$ In fact,$$\frac{c_{n+1}}{c_n}=\frac{d_n}{d_n-1}=\frac{d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}}{d_1\cdot \frac{c_1}{c_1-1}\cdots \frac{c_{n-1}}{c_{n-1}-1}-1} \Rightarrow \frac{c_{n}}{c_{n+1}}=1-\frac{1}{d_1}((1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n-1}}))$$

Together with $d_1(1-\frac{c_{n+1}}{c_{n+2}})=(1-\frac{1}{c_1})\cdots (1-\frac{1}{c_{n}})$, we get $(1-\frac{1}{c_n})(1-\frac{c_n}{c_{n+1}})=1-\frac{c_{n+1}}{c_{n+2}}$. That is $$ \frac{c_{n+1}}{c_{n+2}}-\frac{c_{n}}{c_{n+1}}=\frac{1}{c_{n}}-\frac{1}{c_{n+1}} $$ Hence, for $n\ge 2$ $$ \frac{c_{n}}{c_{n+1}}-\frac{c_2}{c_3}=\frac{1}{c_2}-\frac{1}{c_{n}} $$ which is equivalent to $$\frac{c_{n}}{c_{n+1}}+\frac{1}{c_{n}}=c(constant):=\frac{c_2}{c_3}+\frac{1}{c_2}=\frac{7}{6} \\ (c_1=2,d_1=3;c_2=3,d_2=6;c_3=\frac{18}{5}) $$ Now it is clear that $$\frac{c_{n+1}}{c_{n}}=\frac{c_n}{\frac{7}{6}c_n-1}=\frac{1}{\frac{7}{6}-\frac{1}{c_n}} $$ Well, the problem has been reduced to solve $c_n(=a_n+1)$, and it's you can use the same method to solve $b_n$, and I would like to leave this open to you, but note as I mentioned before if $a_n\to +\infty$, then $c_n\to +\infty$ and $$ \frac{c_{n+1}}{c_{n}}(=\frac{\frac{a_{n+1}}{a_n}+\frac{1}{a_n}}{1+\frac{1}{a_n}})=\frac{1}{\frac{7}{6}-\frac{1}{c_n}} \text{converges to } \frac{6}{7} \text{instead of } 1 $$ this contradicts the last point.

  • $\begingroup$ So this solution shows that $a_n$ converges, but it does not find the value of the limit. $\endgroup$ – Ewan Delanoy Jun 8 '13 at 8:52
  • 1
    $\begingroup$ @EwanDelanoy Yes, but my next answer figured out the limit. $\endgroup$ – Coiacy Jun 8 '13 at 22:15

The answer is $a_n \to 5$ , $b_n \to \infty$.

I'm trying to prove that and I will edit this post if I figure out something.

EDIT: I would write all this in comment instead in answer, but I cannot find how to do it.. maybe I need to have more reputation to do this (low reputation = low privileges:P)

Anyway, I still didn't solved it, but maybe something of that will help you. I will edit it when I think something out.

EDIT: After many transformations and playing with numbers, I think that the limit, for $a<b$, is $$ \frac{ab + 2a +1}{b-a}$$

But still cannot prove it.

(In statement above: $a = a_1 $ , $ b = b_1 $)

  • $\begingroup$ Hi could you tell me how to get the correct formula of the limit? i.e. what kind of software(and related functions) were you using to compute? Thanks! $\endgroup$ – Coiacy May 28 '13 at 0:45

From a previous answer (Coiacy) we know that $a_{n}$ is increasing and $lim_{n\rightarrow\infty} b_{n} = \infty$.

It is easy to prove equalities:

1) $ 1+a_{n+1}=\frac{(1+a_{n})(1+b_{n})}{b_{n}}$;

2) $1+b_{n+1}=\frac{(1+a_{n})(1+b_{n})}{a_{n}}; $

3) $\frac{1}{1+a_{n+1}}-\frac{1}{1+b_{n+1}}= \frac{1}{1+a_{n}}-\frac{1}{1+b_{n}}.$

From 3) it follows that $\frac{1}{1+a_{n}}-\frac{1}{1+b_{n}} = \frac{1}{1+a_{1}}-\frac{1}{1+b_{1}} = \frac{1}{6}$ and $\frac{1}{1+a_{n}}= \frac{1}{6} + \frac{1}{1+b_{n}} > \frac{1}{6}$.

From here we have that $ a_{n}< 5 $ which means that $a_{n}$ is monotone and bounded and $lim_{n\rightarrow\infty} a_{n} = l$ where $l\in (0, 5]$.

Because $lim_{n\rightarrow\infty} b_{n} = \infty$ it follows that $lim_{n\rightarrow\infty}\frac{1}{1+a_{n}}$ $= \frac{1}{6}+lim_{n\rightarrow\infty}\frac{1}{1+b_{n}} = \frac{1}{6}$ and consequently $lim_{n\rightarrow\infty} a_{n} = 5$

  • $\begingroup$ You've made a big discovery and I like it, but it still remians to compute the general form of $a_n$ and $b_n$. $\endgroup$ – Coiacy Jun 11 '13 at 21:18

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