# 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 $$7$$.

• Hint: Substitute $i=n-k$ and use the fact that $(1-\frac{k}{n})^{2n} \uparrow e^{-2x}$ as $n\to\infty$ for $k \geq 0$. This, this, this, this, and this might help. Jul 5, 2020 at 23:43
• These brackets denote the "floor function". Jul 5, 2020 at 23:59
• @WolfgangKais Of course I know that, I meant the question asks "What is $\lfloor \frac{1}{L} \rfloor$"
– user801111
Jul 6, 2020 at 0:45

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.