The simplest solution I can think of starts with the infinite geometric series
S(a) = sum_{n=1}^infty e^(-an) S(a)=∞∑n=1e−an
which converges for a>0a>0 to the well known sum
S(a) = e^(-a)/{1-e^(-a)} = 1/{e^a-1}S(a)=e−a1−e−a=1ea−1
Differentiating both sides with respect to aa leads to
d/{da}(sum_{n=1}^infty e^(-an)) =d/{da} (1/{e^a-1}) impliesdda(∞∑n=1e−an)=dda(1ea−1)⇒
-sum_{n=1}^infty n e^(-an) = -e^a/(e^a-1)^2−∞∑n=1ne−an=−ea(ea−1)2
Substituting a=1a=1 gives us
-sum_{n=1}^infty n e^(-n) = -e/(e-1)^2 implies−∞∑n=1ne−n=−e(e−1)2⇒
-sum_{n=1}^infty n e^(1-n) = -e^2/(e-1)^2−∞∑n=1ne1−n=−e2(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
