First we note that by the distribute property of the sum:
#sum_(k=1)^n (2k+1)/(k!) = 2sum_(k=1)^n k/(k!) + sum_(k=1)^n1/(k!) = 2sum_(k=1)^n 1/((k-1)!) + sum_(k=1)^n1/(k!)#
In the first sum change #k-1= j# so that for #k=1,2,...,n# we have that #j=0,1,...,(n-1)#:
#sum_(k=1)^n (2k+1)/(k!) = 2 sum_(j=0)^(n-1) 1/(j!) + sum_(k=1)^n1/(k!)#
we can add and subtract the terms required to make the sums over the same indeces, that is the term for #j=n# in the first and the term for #k=0# in the second:
#sum_(k=1)^n (2k+1)/(k!) = 3sum_(k=0)^n1/(k!) -1-1/(n!)#
We know that the MacLaurin series for the exponential function is:
#e^x = sum_(k=0)^oo x^k/(k!)#
so for #x=1# and by the definition of the sum of a series:
#e= lim_(n->oo) sum_(k=0)^n 1/(k!)#
So:
#lim_(n->oo) sum_(k=1)^n (2k+1)/(k!) = 3 lim_(n->oo) sum_(k=0)^n1/(k!) -1 - lim_(n->oo) 1/(n!) = 3e-1#
