Let $\{a_n\}$ be a sequence of positive real numbers such that

$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$.

Then the sum of the series $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in...

(A) $(1,2]$, (B) $(2,3]$, (C) $(3,4]$, (D)$(4,5]$.

Solution attempt:

Firstly, we figure out what $\frac{a_{n+1}}{a_n}$ is going to look like. We get, from the recursive formula, $\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$ (remembering the fact that $a_n>0$, the other root is rejected).

We know that, if $\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$, then $\lim a_n \to \infty$.

Further, $(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$. (Again, the other root is rejected due to the same reason).

Hence, $(a_n)$ increases monotonically.

Therefore, the largest value of $\frac{a_{n+1}}{a_n}$ is approximately $1+\sqrt{1+\frac{1}{1}} \approx 2.15$

Now, the sum can be approximated as $\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$ (In actuality, $\mathbb{sum}< 1.3$).

So, option $(A)$ is the correct answer.

Is the procedure correct?

I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie.

Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem?

  • $\begingroup$ Out of curiosity, what's the source of this problem ? $\endgroup$ – Gabriel Romon May 19 '19 at 10:16
  • $\begingroup$ @GabrielRomon It was asked in an entrance exam for Master's degree admission in India this year (JAM 2019). $\endgroup$ – Subhasis Biswas May 19 '19 at 10:18
  • $\begingroup$ Hmm, your approach is indeed quite nice, but even if asymptotically $a_n\sim\alpha^n$ and the sum effectively $\frac\alpha{3-\alpha}$, the first terms of the series may as well shift the result in another interval. How do you bound the partial sum $\sum\limits_{n=1}^{n_0} \frac{a_n}{3^n}$ up the a certain $n_0$ so that subsequent terms are small enough and we can switch to asymptotic behaviour ? $\endgroup$ – zwim May 19 '19 at 10:48
  • $\begingroup$ This is the part where I used the monotone property. The common ratio can never exceed $2.15$, no matter what. Because, after the first term of the sequence, $1/a_{n}^2 <1$, Resulting in $a_{n+1}/a_n <2.15$ $\endgroup$ – Subhasis Biswas May 19 '19 at 10:50
  • 1
    $\begingroup$ If you recall the proof of the ratio test, then yes, you applied the right strategy. $\endgroup$ – rtybase May 19 '19 at 10:58

Following your calculations and according to the ratio test $$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_n}{3^n}}=\frac{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$ thus $$\sum\limits_{n=1}\frac{a_n}{3^n}< \infty$$ Now, applying the same technique from the proof on the ratio test $$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^3}+\cdots+\frac{a_n}{3^n}+\cdots=\\ \frac{1}{3}+\frac{a_2}{a_1}\cdot\frac{a_1}{3^2}+\frac{a_3}{a_2}\cdot\frac{a_2}{3^3}+\cdots+\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}+2\cdot\frac{1}{3^2}+2\cdot\frac{a_2}{3^3}+\cdots+2\cdot\frac{a_{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+2.15\cdot\frac{1}{3^2}+2.15\cdot\frac{a_2}{3^3}+\cdots+2.15\cdot\frac{a_{n-1}}{3^n}+\cdots$$ and repeating this $$\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\cdots+\frac{2^{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+\frac{2.15}{3^2}+\frac{2.15^2}{3^3}+\cdots+\frac{2.15^{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}\cdot\left(1+\frac{2}{3}+\frac{2^2}{3^2}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots\right)< S<\\ \frac{1}{3}\cdot\left(1+\frac{2.15}{3}+\frac{2.15^2}{3^2}+\cdots+\frac{2.15^{n-1}}{3^{n-1}}+\cdots\right)$$ or $$\color{red}{1}=\frac{\frac{1}{3}}{1-\frac{2}{3}}<\color{red}{S}<\frac{\frac{1}{3}}{1-\frac{2.15}{3}}=\frac{1}{3-2.15}<\color{red}{2}$$

This kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask something similar for a master degree entrance test.

  • 3
    $\begingroup$ Truly a nice approach. The upper bound is basically the same. I missed the lower bound though :( I like your final note. "So, it makes sense to ask something similar for a master degree entrance test" $\endgroup$ – Subhasis Biswas May 19 '19 at 11:33
  • $\begingroup$ Nice and clean :) $\endgroup$ – A learner Apr 23 at 15:44

$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$ $a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$ $1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$ $a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$ $\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$ $\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$ $\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$ $\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$


I would rather argue that $\sqrt{a_n(a_n+1)}$ lies inside $[a_n,a_n+1]$, so that $b_n\leq a_n \leq c_n$ where $b_n=2b_n$ and $c_n=2c_n+1$ with $b_1=c_1=1$.

Closed forms for $b_n$ and $c_n$ are easily derived as $b_n=2^{n-1}$ and $c_n=2^n-1$, so that $$1\leq \sum_1^{\infty}\frac{a_n}{3^n} \leq 2-\frac 12$$

This inequality can be refined by only summing from $n$ larger than some constant.

  • $\begingroup$ Now, for the last part of my question, is there any general approach for these type of problems? $\endgroup$ – Subhasis Biswas May 19 '19 at 11:07
  • $\begingroup$ A better one indeed! In fact, we don't need any better bound than this. Although my procedure gives off a slightly better value, it is not worthwhile here. $\endgroup$ – Subhasis Biswas May 19 '19 at 11:08

(Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.)


  • After calculating the first few terms, the ratio of the terms is very similar. If we estimate $ a_n \approx a_1 r^{n-1}$, this suggests we should solve for $ \frac{ 1/3} { 1 - (r_1/3) } = 1$ and $ \frac{1/3} { 1 - (r_2/3)} = 2$ to give us an idea of how to bound the sequence. This gives us $ r_1 = 2, r_2 = 5/2$, so we want to show that $ 2 a_n < a_{n+1} < \frac{5}{2} a_n$ (with some flexibility if this doesn't immediately work out).
  • Show that $ a_{n+1} = \frac{ 2a_n + \sqrt{ 4a_n^2 + 4a_n } } { 2} $. In particular, reject the negative root.
  • Show that $ a_n \geq 1$.
  • Show that $ 2 a_n < a_{n+1} < \frac{5}{2} a_n$.
  • Hence, show that $1=\frac{ 1/3 } { 1 - 2/3} < \sum \frac{a_n}{3^n} < \frac{ 1 / 3 } { 1 - 5/6}=2 $


  • The LHS is true by calculating the first 5 terms.
  • Of course, we can't prove the RHS just by calculating enough terms.
  • In fact, the bounding inequality $ 2a_n < a_{n+1} < 2a_n + \frac{1}{2}$, so the ratio $ a_{n+1} / a_n$ is very close to 2, esp at (slightly) larger values of $n$.
  • Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25.

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