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Check the convergence of $\sum_{n=1}^{\infty} ( e - (1+ \frac{1}{n})^{n})$

$$\lim_{n \to \infty} \left(e-\left(1+ \frac{1}{n}\right)^{n}\right) = 0$$

So, divergence test fails. I couldn't find any suitable test to check the convergence of this series.

Any suggestion $?$

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We have that

$$\left(1+ \frac{1}{n}\right)^{n}=e^{n\log \left(1+ \frac{1}{n}\right)}=e^{1-\frac1{2n}+O\left(\frac1{n^2}\right)}=e\left(1-\frac1{2n}+O\left(\frac1{n^2}\right)\right)$$

therefore

$$e - \left(1+ \frac{1}{n}\right)^{n}=\frac e{2n}+O\left(\frac1{n^2}\right)$$

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    $\begingroup$ The same question (with identical answers) has been asked multiple times. Even you answered the question already. $\endgroup$ – Martin R Dec 4 '19 at 14:32
  • $\begingroup$ @MartinR I see now. I didn't remember it. I see that I used another method that time. $\endgroup$ – user Dec 4 '19 at 14:46
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    $\begingroup$ Do don't have to remember it. There is an easy-to-use search engine to find obvious duplicates. Your method is used in all the referenced duplicate targets. I do not see that it adds something substantially new. $\endgroup$ – Martin R Dec 4 '19 at 14:48
  • $\begingroup$ @MartinR Some users have upvoted it then it means that it was useful to someone in the community. Someone has downvoted it, which is always wrong in my opinion for correct answers. $\endgroup$ – user Dec 4 '19 at 14:57

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