A) this series is a geometric series: a=1;q=e−1a=1;q=e^{-1}a=1;q=e−1 - false
B) For the geometric series Sn=a(1−qn)1−qS_n=\frac{a(1-q^n)}{1-q}Sn=1−qa(1−qn) therefore Sn=1−e−n1−e−1S_n=\frac{1-e^{-n}}{1-e^{-1}}Sn=1−e−11−e−n
C) limn→∞Sn=11−e−1=ee−1\lim\limits_{n\to\infty}S_n=\frac{1}{1-e^{-1}}=\frac{e}{e-1}n→∞limSn=1−e−11=e−1e - true
D) this series is not a telescoping series
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