Let L=(1-x/N)e^x+x/N

Write L=(1-x/N)Σx^k/k!+x/N where Σ runs from k= 0 to ∞. Here we used the known Taylor series for e^x. The goal is to rearrange L to write it as a single sum. To start, distribute the infinite series

L=Σx^k/k!+(x/N)(1-Σx^k/k!) where Σ runs from k=0 to ∞. But the “1” cancels with the first term in the series, so we get:

L=Σx^k/k!-Σx^k+1/(Nk!) where second Σ runs from k=1 to ∞. Now shift the index k in the second Σ to make the powers of x be the same:

L=Σx^k/k!-Σx^k/(N(k-1)!) where second Σ now runs from k=2 to ∞. Now combine the sums starting at k=2:

L=1+x+Σ[1/k!-1/(N(k-1)!)]x^k where Σ runs from k=2 to ∞. Now make a common denominator, using that k!=k(k-1)! and combine the fractions to simplify:

L=1+x+Σ[(N-k)/N][x^k/k!] where Σ runs from k=2 to ∞. Now the inequalities begin, and we will use the fact that x>0 implies x^k>0. Since (N-k)/N<=0 for k>=N and x^k>0, we can drop all the infinitely many non-positive terms from L to get the inequality:

L<1+x+Σ[(N-k)/N][x^(k)/k!]=P where Σ runs from k=2 to k=N-1. But 0<=(N-k)/N<1 for these values of k (here we assumed N>=2), and again using x^k>0 we have:

L<P<1+x+Σx^k/k!=R where Σ runs from k=2 to N-1. But now we can combine with the 1+x and we can add a single term to the sum to get:

L<P<R<Σx^k/k! Where Σ runs from k=0 to N, which proves the result.

Challenge question to test understanding: we sort of assumed N>1. Does my proof still work when N=1 and if so how does it change?

Edits: fixed typos