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Let the sequence $\{x_n, n=1,2,...\}$ be defined as follows: Let $x_2=x_1=1$ and for $n>2$ let $$x_{n+1}=x_n-\frac{1}{n}x_{n-1}.$$

This sequence, generated by the recursion above, tends to zero extremely fast. My guess is that

$$\sum_{i=1}^{\infty}x_i=e.$$

I came across this problem here on MSE while working on an infinite Markov chain related problem. The state transition probability matrix was given and the task was to find the stationary probabilities. The stationary probabilities could be generated by the recursion above but with an unknown $c=x_1=x_2$. I did some experiments with $c$ and the results lead me believe that $$c=e.$$

I would be surprised if this recursion was not known to somebody around here.

Community
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zoli
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3 Answers3

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It can be proved by induction that $x_n = \frac{1}{(n-1)!}$

$x_1$ and $x_2$ satisfy this, and by doing induction,

$$x_{m+1} = x_{m}-\frac{1}{m}x_{m-1} = \frac{1}{(m-1)!} - \frac{1}{m}\frac{1}{(m-2)!} = \frac{1}{m!} $$

Thus you series reduces to the ordinary series $\sum_{n=0}^{\infty}\frac{1}{n!}$ for $e$

mastrok
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5

The recursion is satisfied by $$x_n=\frac{1}{\Gamma (n)}$$ So, $$\sum_{i=1}^p x_i=\frac{e\, p \,\Gamma (p,1)}{\Gamma (p+1)}$$ where appears the incomplete gamma function and the limit is effectively $e$.

Claude Leibovici
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3

The first terms of the sequence are $$1,1,\frac12,\frac16,\frac1{24},\frac1{120}\cdots$$which you recognize without surprise.

Indeed, with $x_n=\dfrac1{(n-1)!}$ you have

$$x_n=nx_{n+1},$$ $$x_{n-1}=n(n-1)x_{n+1},$$ hence $$x_n-\frac nn(n-1)x_{n-1}=x_{n+1}.$$

Nothing new, sorry.