Cool Infinite Sum!

Probability Level 4

Let $S$ be the infinite sum given by $S = \sum_{n=0}^{\infty} \frac{a_n}{10^{2n}}$ where $\langle a_n \rangle_{n \geq 0}$ is a sequence defined by $a_0 = a_1 = 1$ and $a_j = 20a_{j-1} - 108a_{j-2}$ for $j \geq 2$ . If $S$ can be expressed in the form of $\dfrac{a}{b}$ , where $a$ and $b$ are coprime positive integers, then what is $a$ ?

2020 2023 2018 2025

3 solutions

Mark Hennings
May 22, 2018

Solving the recurrence relation, we obtain $a_n \; = \; \frac{2\sqrt{2} + 9i}{4\sqrt{2}}(10 + 2i\sqrt{2})^n + \frac{2\sqrt{2} - 9i}{4\sqrt{2}}(10 - 2i\sqrt{2})^n$ for all $n$ , and so the sum is $\begin{aligned} \sum_{n=0}^\infty \frac{a_n}{10^{2n}} & = \; \frac{2\sqrt{2} + 9i}{4\sqrt{2}}\sum_{n=0}^\infty \left(\frac{10 + 2i\sqrt{2}}{100}\right)^n + \frac{2\sqrt{2} - 9i}{4\sqrt{2}}\sum_{n=0}^\infty \left(\frac{10 - 2i\sqrt{2}}{100}\right)^n \\ & = \; \frac{2\sqrt{2} + 9i}{4\sqrt{2}} \times \frac{1}{1 - \frac{10 + 2i\sqrt{2}}{100}} + \frac{2\sqrt{2} - 9i}{4\sqrt{2}} \times \frac{1}{1 - \frac{10 - 2i\sqrt{2}}{100}} \; = \; \frac{2025}{2027} \end{aligned}$ making the answer $\boxed{2025}$ .

FYI: the denominator of the parenthetical quantities in your second line should be $100$ instead of $10.$

Patrick Corn - 3 years ago

So true...

Mark Hennings - 3 years ago

Patrick Corn
May 22, 2018

First we show that the sum converges, using the following lemma:

Lemma: $|a_n| \le 50^n$ for all $n.$

Proof: By induction. It's true for $n=0,1,$ and then supposing it's true for $0,1,\ldots,k-1,$ we have $|a_k| \le 20|a_{k-1}| + 108|a_{k-2}| \le 20 \cdot 50^{k-1} + 108 \cdot 50^{k-2} = 50^k \left( \frac25 + \frac{108}{2500} \right) < 50^k.$

So the sum converges because the absolute value of the $n$ th term is $\le (1/2)^n.$

To evaluate the sum, write $\begin{aligned} \sum_{n=2}^\infty \frac{a_n-20a_{n-1}+108a_{n-2}}{10^{2n}} &= 0 \\ \sum_{n=2}^\infty \frac{a_n}{10^{2n}} - 20\sum_{n=2}^\infty \frac{a_{n-1}}{10^{2n}} + 108\sum_{n=2}^\infty \frac{a_{n-2}}{10^{2n}} &= 0 \\ \left(S - 1 - \frac1{100}\right) - \frac15 \sum_{n=2}^\infty \frac{a_{n-1}}{10^{2(n-1)}} + \frac{108}{10000}\sum_{n=2}^\infty \frac{a_{n-2}}{10^{2(n-2)}} &= 0 \\ \left(S - 1 - \frac1{100}\right) - \frac15 (S - 1) + \frac{108}{10000} S &= 0 \\ S\left( 1-\frac15 + \frac{108}{10000} \right) &= 1-\frac15 + \frac1{100} \\ S &= \frac{10000-2000+100}{10000-2000+108} = \frac{2025}{2027}, \end{aligned}$ so $a=\fbox{2025}.$

Oh wow, I'm not familiar with this technique of finding the sum. Thanks! +1

Just curious, why do we have to prove that the sum converges? I feel like your first half of the solution is not necessary. I mean, if we have found the sum to be finite, doesn't it automatically implies that the sum converges? Or is there a case where the sum converges "through other means" and we just want to make sure it converges "conventionally"?

EDIT: Oh wait, nevermind. I got it. You have defined the sum to be $S$ in your latter part of your solution. To do so, we need to prove that $S$ is finite first. Otherwise, there might be a case of $\infty - \infty$ .

Pi Han Goh - 3 years ago

Alapan Das
Mar 26, 2019

I shall right an easy explanation. Say, (a j)/10^2j=P j. So, S=P 1+P 2+..... Now ,you easily can find the following relation P j=0.2P j-1-0.0108P j-2 ,P j-1=0.2P j-2 -0.0108P j-3... ... ,... ,P 2=0.2P 1 - 0.0108P 0 Now add L.H.S and R.H.S. P 2+P 3+...=0.1892(P 1+P 2+...) -0.0108. Add P 1+P_0 to both sides. Then it becomes S=0.1892(S-1)+1.01-0.0108 S=0.1892S+0.81 S=2025/2027. So, a=2025

Cool Infinite Sum!

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