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The sequence $x_1,x_2,x_3,\cdots$ is defined by $x_1=2$ and $x_{k+1}=x_k^2-x_k+1$ for all $k \ge 1$.

Find $\sum_{k=1}^\infty \cfrac{1}{x_k} $

By experimenting ,I was able to prove by induction that $$\sum_{k=1}^j \cfrac{1}{x_k}=\cfrac{x_{j+1} -2}{x_{j+1}-1}$$

But now I am quite unsure on how to sum over infinity,do I just treat it as a number and let $j=\infty$ ?This seems a little bit fishy to do...

Can someone help me ?

Mr. Y
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    I have not checked your formula but if you compute the first values of your sequence, you will see that it seems to tend very fast to $+\infty$ and consequently, $(x_{j+1}-2)/(x_{j+1}-1) = 1-1/(x_{j+1}-1)$ tends to... – user37238 Jan 19 '16 at 09:07
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    And in fact you can easily prove by induction that the sequence is strictly increasing. I did verify your formula. – Brian M. Scott Jan 19 '16 at 09:25

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By definition, $$\sum_{k=1}^\infty a_k = \lim_{j\to\infty}\sum_{k=1}^j a_k$$

meaning that in your case, if your formula is correct, you will have $$\sum_{k=1}^\infty = \lim_{j\to\infty}\frac{x_{j+1}-2}{x_{j+1}-1} = \lim_{j\to\infty}\frac{1 - \frac{2}{x_{j+1}}}{1 - \frac{1}{x_{j+1}}}$$

Which is equal to $1$ because $x_j\to\infty$ as $j\to\infty.$

5xum
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