Finding limit using Riemann integral problem

How do I evaluate the following limits ?

$$\lim_{n\to +\infty}\left(\frac{1}{n^2}\sum_{k=1}^n\left(2k-1\right) \sqrt[n]{e^{2k-1}}\right)$$

In this case I'm confused about function to integrate: $f\left(x\right)=\:x\cdot e^x ?$ Please help and explain to understand how I find the function to integrate...

$$\lim _{n\to \infty }\frac{1}{n^2}\cdot \sqrt[n]{\left(n^2+1\right)\cdot \left(n^2+2\right)\cdot \left(n^2+5\right)\cdot \:\:\:...\:\:\:\cdot \left(2n^2-2n+2\right)}$$

$$e^{-1/n}\left(\frac{1}{n}\sum_{k = 1}^n \frac{2k}{n} e^{2k/n}\right) - \frac{e^{-1/n}}{n}\left(\frac{1}{n}\sum_{k = 1}^n e^{2k/n}\right).$$

In the first term, the expression in parentheses is a Riemann sum of $2xe^{2x}$ over $[0,1]$. So it converges to

$$\int_0^1 2xe^{2x}\, dx = \frac{1}{2}\int_0^2 ue^u\, du = \frac{1}{2}(u - 1)e^u\bigg|_{u = 0}^2 = \frac{1}{2}(1 + e^2).$$

Since $e^{-1/n} \to 1$, it follows that the first term converges to $\frac{1}{2}(1 + e^2)$. In the second term, the expression in parentheses is a Riemann sum of $e^{2x}$ over $[0,1]$. So it converges (to $\int_0^1 e^{2x}\, dx$). Since $e^{-1/n} \to 1$ and $1/n \to 0$, it follows that the second term converges to $0$. Therefore, your limit is $\frac{1}{2}(1 + e^2)$.

• But that is not equal with this $\lim _{n\to \infty }\sum _{k=1}^n\left(\frac{\left(2k-1\right)}{n}\cdot e^{\frac{\left(2k-1\right)}{n}}\right)$ ? Feb 27, 2015 at 8:11
• In your first formula, I think it is $2k/n-1/n$, (and so you have to split the formula in two parts) Feb 27, 2015 at 8:12
• @Andrei sorry, misread. I will make changes.
– kobe
Feb 27, 2015 at 8:12
• @Andrei I've made the appropriate edits, assuming I haven't misread anything. :)
– kobe
Feb 27, 2015 at 8:26
• but I don't understand why is $$2x\cdot e^{2x}$$ and not $$x\cdot e^x$$ ? because I think $$\frac{\left(2k-1\right)}{n}$$ equal with x in our function Feb 27, 2015 at 8:37