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I would like to find the domain of convergence of the series $\sum\limits_{n=1}^{\infty} \left(e - \left(1+\dfrac{1}{n}\right)^n\right)^{2x}$.

In fact, I knew that $\lim \left(e - \left(1+\dfrac{1}{n}\right)^n\right) = 0$ and then this series is divergent when $x \leq 0$.

Do you have a solution to this problem when $x> 0$?

Thank you so much!

Claudio
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Mariod
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    It might help to consider the function $$f(z)=e-e^{\ln(1+z)/z}.$$ Note that $f$ is analytic around $z=0$. Use this to get a suitable bound for your series. – Severin Schraven Jun 17 '24 at 18:11
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    With my hint you get $$e-(1+1/n)^n=f(1/n)=e-e^{1-1/(2n)+O(1/n^2)}=e/(2n)+O(1/n^2)$$ which should allow you to conclude. – Severin Schraven Jun 17 '24 at 18:27

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May be you can consider some bounded inequalities like $e-(1+\frac{1}{n})^n <\frac{3}{n}$.

Five_Seven
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We can rewrite $$ e - \left( 1+\frac{1}{n} \right)^n = e - e^{\ln(1+1/n)/(1/n)}. $$ For $y$ small we have using Taylor series $$ e-e^{\ln(1+y)/y} = e -e^{1-\frac{1}{2}y+O(y^2)} = e - (e-e\frac{1}{2}y+O(y^2)) = \frac{e}{2}y+O(y^2). $$ With $y=1/n$ we get $$ e - \left( 1+\frac{1}{n} \right)^n = e - e^{\ln(1+1/n)/(1/n)} = \frac{e}{2} \frac{1}{n}+O(1/n^2). $$ Thus, for $n$ large enough, we get $$ \frac{1}{n} \leq e - \left( 1+\frac{1}{n} \right)^n \leq \frac{2}{n}. $$ By the comparison test, we get that $$ \sum_{n=1}^\infty \left(e - \left( 1+\frac{1}{n} \right)^n \right)^{2x} $$ converges if and only if $$ \sum_{n=1}^\infty \frac{1}{n^{2x}}$$ converges. By the $p$-test, we get that the later converges if and only if $2x>1$. Thus, our series converges for $x>1/2$ and diverges otherwise (if one is pedantic, then one also needs this Show that $\left(1+\dfrac{1}{n}\right)^n$ is monotonically increasing to ensure that can raise our terms to a non-integer power).

Severin Schraven
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In the very same spirit as @SeverinSchraven's answer but in a lighter formulation, using the symbol $\sim$ of asymptotic equivalence:

as $n\to\infty$

$$u_n:=n\ln\left(1+\frac1n\right)-1\sim-\frac1{2n}\to0$$ hence $$v_n:=e-\left(1+\frac1n\right)^n=e\left(1-e^{u_n}\right)\sim-eu_n\sim\frac e{2n}.$$ Therefore, $${v_n}^{2x}\sim\left(\frac e2\right)^{2x}\frac1{n^{2x}},$$ so that $\sum{v_n}^{2x}$ converges iff $2x>1$.

Anne Bauval
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