Question:exp function

Question:exp function

Maple
hi, i am trying to show that the sum of (1/2 + 2/3 + 3/4 + 4/5 + ... +n*(n+1)+ ...) can be approximated by the following: sum(1/(exp(1))^(1/n), n = 1 .. infinity). e.g. > sum(1/exp(1/n), n = 1 .. 10); print(1/exp(1)+1/exp(1/2)+1/exp(1/3)+1/exp(1/4)+1/exp(1/5)+1/exp(1/6)+1/exp(1/7)+1/exp(1/8)+1/exp(1/9)+1/exp(1/10)); 1 1 1 1 1 1 1 1 1 ------ + ------ + ------ + ------ + ------ + ------ + ------ + ------ + ------ exp(1) /1\ /1\ /1\ /1\ /1\ /1\ /1\ /1\ exp|-| exp|-| exp|-| exp|-| exp|-| exp|-| exp|-| exp|-| \2/ \3/ \4/ \5/ \6/ \7/ \8/ \9/ 1 + ------- /1 \ exp|--| \10/ > evalf(1/exp(1)+1/exp(1/2)+1/exp(1/3)+1/exp(1/4)+1/exp(1/5)+1/exp(1/6)+1/exp(1/7)+1/exp(1/8)+1/exp(1/9)+1/exp(1/10)); print(7.684006210); 7.684006210 > sum(n/(n+1), n = 1 .. 10); print(221209/27720); 221209 ------ 27720 > evalf(221209/27720); print(7.980122655); 7.980122655 7.684006210 vs 7.980122655 However, for some reason Maple will not show that the infinite sum of this function tends to plus infinity. Instead I just get the formula that I put in the first place. > limit(sum(1/exp(1/n), n = 1 .. infinity), n = infinity); print(sum(1/exp(1/n), n = 1 .. infinity)); infinity ----- \ ) 1 / ------ ----- /1\ n = 1 exp|-| \n/ Would appreciate advice on why this is happening. many thanks. antonio
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