There's something hidden in it

Calculus Level 5

${a}_{n+1} +1 = (n+1){a}_{n}$

Define an sequence $\lbrace a_{n} \rbrace_{n=0}$ by the iteration above. For what value of ${a}_{0}$ will the iteration converge to some finite value?

Report your answer as first four digits of ${a}_{0}$ (without rounding off). For example, if your answer is $3.141592653789 \ldots$ then enter 3.141.

Interestingly in the complete number line there exists only one possible value of ${a}_{0}$ , for which this iteration doesn't diverge (that's the interesting aspect of it)

The sequence diverge to positive infinity for any starting value larger than $a_0$ , and the sequence diverge to negative for any starting value smaller than $a_0$ .

The answer is 1.718.

2 solutions

Jake Lai
Mar 15, 2015

This isn't a very formal solution, so I apologise.

Starting off with $a_{0}$ :

$\begin{array}{c}\ a_{0} = a_{0} \\ a_{1} = 1!a_{0}-1 \\ a_{2} = 2!a_{0}-\frac{2!}{1!}-\frac{2!}{2!} \\ a_{3} = 3!a_{0}-\frac{3!}{1!}-\frac{3!}{2!}-\frac{3!}{3!} \end{array}$

Going on, we see that

$a_{n} = n! \times a_{0}-\frac{n!}{1!}-\frac{n!}{2!}-\ldots-\frac{n!}{n!} = n! \left( a_{0} - \sum_{k=1}^{n} \frac{1}{k!} \right)$

The only way for $a_{n}$ to converge as $n \rightarrow \infty$ is for

$a_{0} = \sum_{k=1}^{n} \frac{1}{k!}$

Since our RHS is $e-1$ by the Taylor series of $e^{x}$ , $a_{0} = e-1 \approx \boxed{1.718}$ . The sequence thus converges to $0$ .

Yep! Correct! BTW, even ore fancy notation would be to write sum of permutations(which can be proven by using Gamma function) too like ${a}_{n} = n!{a}_{0} - e(\Gamma(n+1,1) - n!)$

${a}_{n} = n!({a}_{0} +1) - e\Gamma(n+1,1)$

This is because this will be the general term and not the above summation one. *From here too, $e-1$ can be easily found.

Kartik Sharma - 6 years, 2 months ago

Good job! This is right.

Raghav Vaidyanathan - 6 years, 3 months ago

Same way! awesome question though, looks hard at first sight ,

but it is necessary to first show that if the sequence does converge, its gonna be at 0, which can ofcourse be easily shown by using $a_{n-1} = a_{n}$

only then, we can claim that the only value is 1.718

Mvs Saketh - 6 years, 2 months ago

Neelesh Vij
Sep 14, 2016

The given relation is:

$a_{n+1} +1 = (n+1)a_{n}$

Divide by $(n+1)!$

$\dfrac{a_{n+1}}{(n+1)!} + \dfrac{1}{(n+1)!} = \dfrac{a_{n}}{n!}$

Let $\dfrac{a_{n}}{n!} = T_{n}$

So we get

$T_{n+1} - T_{n} = - \dfrac{1}{(n+1)!}$

Now as $n \to \infty$ $a_{n+1} = a_{n}$ so $a_{\infty} = 0$

Solving the above sequence we get

$\displaystyle a_0 = \sum_{n=0}^{\infty} \dfrac{1}{n!}$

$\therefore \boxed{a_{0} = e-1}$

I think your summation (second to last) should start at $1$. Other than that, great solution!

Keller VandeBogert - 4 years, 7 months ago

There's something hidden in it

The answer is 1.718.

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