If $L=\lim_{n \to \infty} \sum_{i=1}^{n-1} {\left(\frac{i}{n}\right)}^{2n}$ what is $\lfloor \frac{1}{L} \rfloor$

Question

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 Riemann sum? If not what do I do. Answer is $7$.

Answer 1

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.