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For which real $a$ does $ \sum_{n=1}^{\infty} \left\{ e-\left(1+\frac{1}{n}\right)^{n+a} \right\}$ converge or diverge?

This is a generalization of $\sum_{n=1}^{\infty} \left\{ e-(1+\frac{1}{n})^n \right\}$. is this converge or diverge where it is shown that the sum diverges for $a=0$.

My conjecture is that the sum converges for $a = \frac12$ and diverges for $a \ne \frac12$.

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I think you are correct.

Note that

$$\begin{align}\log \left(1 + \frac{1}{n} \right)^{n +a} &= \frac{n+a}{n+1/2}(n + 1/2)\log\left(1 + \frac{1}{n} \right)\\&=\frac{n+a}{n+1/2}(n + 1/2)\left(\frac{1}{n} - \frac{1}{2n^2} +\mathcal{O}\left(\frac{1}{n^3} \right) \right) \\ &= \frac{n+a}{n+1/2}\left(1 + \mathcal{O}\left(\frac{1}{n^2} \right) \right) \\ &= \left(1 + \frac{a - 1/2}{n + 1/2} \right)\left(1 + \mathcal{O}\left(\frac{1}{n^2} \right) \right)\end{align}$$

We have convergence if $a = 1/2$ and divergence otherwise, since

$$e- \left(1 + \frac{1}{n} \right)^{n +a}= e - \exp \left(1 + \frac{a - 1/2}{n + 1/2} + \mathcal{O}\left(\frac{1}{n^2}\right)\right) \\ = e\frac{ 1/2-a}{n + 1/2} + \mathcal{O}\left(\frac{1}{n^2} \right)$$

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Your conjecture is correct, since $$(n+\tau)\log\left(1+\frac{1}{n}\right) = 1+\frac{\tau-\frac{1}{2}}{n}+\frac{\frac{\tau}{2}-\frac{1}{3}+o(1)}{n^2} $$ implies, by exponentiation, $$ \left(1+\frac{1}{n}\right)^{n+\tau}=e\left[1+\frac{\tau-\frac{1}{2}}{n}+\frac{\frac{11}{24}-\tau+\frac{\tau^2}{2}+o(1)}{n^2}\right].$$ Roughly $\sum_{n\geq 1}\left[e-\left(1+\frac{1}{n}\right)^{n+\frac{1}{2}}\right]\approx -\frac{2}{9}$.

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