Find the sum of the infinite series #-ne^(1-n)# from #n=1# to #oo#?

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Answer 1 · 2018-03-05T04:49:51

# -e^2/(e-1)^2#

The simplest solution I can think of starts with the infinite geometric series

#S(a) = sum_{n=1}^infty e^(-an) #

which converges for #a>0# to the well known sum

#S(a) = e^(-a)/{1-e^(-a)} = 1/{e^a-1}#

Differentiating both sides with respect to #a# leads to

#d/{da}(sum_{n=1}^infty e^(-an)) =d/{da} (1/{e^a-1}) implies#
# -sum_{n=1}^infty n e^(-an) = -e^a/(e^a-1)^2#

Substituting #a=1# gives us

# -sum_{n=1}^infty n e^(-n) = -e/(e-1)^2 implies#
# -sum_{n=1}^infty n e^(1-n) = -e^2/(e-1)^2#

Alternative

We can evaluate the sum by a purely algebraic approach as well:

#-S = 1+ 2/e+3/e^2+ 4/e^3+... #

#-1/e S = 1/e +2/e^2+3/e^3 +... #

Subtracting, we get

#-(1-1/e) S = 1 + 1/e+1/e^2+1/e^3+... = 1/(1-1/e)#

This immediately gives the answer