If $L=\lim_{n \to \infty} \sum_{i=1}^{n-1} {\left(\frac{i}{n}\right)}^{2n}$ what is $\lfloor \frac{1}{L} \rfloor$ If $$L=\lim_{n \to \infty} \sum_{i=1}^{n-1} {\left(\frac{i}{n}\right)}^{2n}$$What is $\lfloor \frac{1}{L} \rfloor$
I really am confuse.  Can this be converted into riemman sum?  If not what do I do.  Answer is $7$.
 A: The limit is
$$
L = \frac{1}{e^2-1}
$$
and so $1/L = e^2 - 1 \approx 6.4$, so the result should be 6.
To see this, note that the last element of the sum gives
$$
\left ( \frac{n-1}{n} \right )^{2n} \to e^{-2},
$$
the two last ones give
$$
\left ( \frac{n-2}{n} \right )^{2n} + \left ( \frac{n-1}{n} \right )^{2n} \to e^{-4} + e^{-2}
$$
and so on. Then
$$
e^{-2} + e^{-4} + \cdots = \frac{1}{e^2-1} = L.
$$
Of course, this is not a proof, because there is a whole bunch of inversion of limits that occur. But the argument definitely shows that you have
$$
\liminf S_n \geq L,
$$
where I denote the sum by $S_n$. But what about the limsup? To see this, you can indeed use a Riemann sum, or rather compare directly with an integral. Draw a picture to see that, for any fixed $1 \leq k \leq n$
$$
\frac1n \sum_{i=0}^{n-k} \left (\frac{i}{n} \right)^{2n} \leq \int_0^{1-(k-1)/n} x^{2n} \: \mathrm{d}x = \frac{1}{2n+1} \left ( 1- \frac{k-1}{n} \right )^{2n} ,
$$
so
$$
\limsup \sum_{i=0}^{n-k} \left (\frac{i}{n} \right)^{2n} \leq \frac12 e^{-2(k-1)}.
$$
You therefore get that, by splitting the sum in the $k - 1$ last elements and the rest,
$$
\limsup S_n \leq e^{-2} + e^{-4} + \cdots + e^{-2k+2} + \frac12 e^{-2k+2},
$$
so taking $k \to + \infty$ gives
$$
\limsup S_n \leq e^{-2} + e^{-4} + \cdots = \frac{1}{e^2 - 1} = L,
$$
and we are done.
