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∑n=1∞(1−1e)nn= ?\large\displaystyle\sum_{n=1}^{\infty}\frac{\left(1-\frac{1}{e}\right)^n}{n}=\ ?n=1∑∞n(1−e1)n= ?
With e=∑m=0∞1m!≈2.71828\displaystyle e = \sum_{m=0}^\infty \frac1{m!} \approx 2.71828 e=m=0∑∞m!1≈2.71828.
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