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∑n=1∞(−1)n−1n8n!=ke\large \displaystyle \sum_{n=1}^{\infty} {(-1)}^{n-1} \dfrac{n^8}{n!} = \dfrac ken=1∑∞(−1)n−1n!n8=ek
What is ∣k∣| k | ∣k∣?
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