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# determine whelther the following series converges. find its sum.?

series from n=1 to infinity of ( ( 2 / e^n - 2 / e^n+1 ) )

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- kbLv 78 years agoFavorite Answer
Σ(n = 1 to ∞) [2/e^n - 2/e^(n+1)]

= 2 * lim(k→∞) Σ(n = 1 to k) [e^(-n) - e^(-(n+1))]

= 2 * lim(k→∞) [(e^(-1) - e^(-2)) + (e^(-2) - e^(-3)) + ... + (e^(-k) - e^(-(k+1)))]

= 2* lim(k→∞) (e^(-1) - e^(-(k+1)), since all other terms cancel in pairs

= 2 [e^(-1) - 0]

= 2e^(-1), convergent.

I hope this helps!

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