We have #ainNN and L(a)=lim_(n->oo)1/nint_0^1|__n e^(ax)__|dx.#Which is the solution of inequation #L(a)<=e#?

1 Answer
Oct 17, 2017

#a le 0.402028#

Explanation:

Considering that

#lim_(n->oo) floor(gamma/n)/n = gamma# we have

#lim_(n->oo)1/nint_0^1floor(n e^(ax))dx = int_0^1lim_(n->oo)floor(n e^a)/n dx = int_0^1e^(ax) dx = e^a/a-1#

or

#L(a) = e^a/a-1# so the condition

#L(a) le e rArr a le-W(-1/(e+1)) approx 0.402028#

Here #W# is the so called Lambert function

https://en.wikipedia.org/wiki/Lambert_W_function