Find $$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \left(e^{(1+\frac{k}{n})^2} - \frac{3e^{(1 + \frac{3k}{n})}}{2\sqrt{1 + \frac{3k}{n}}}\right)$$


A: 1
B: $\frac{1}{2}$
C: $\frac{1}{4}$
D: 0

My attempt is:

The question is directly equal to $$\int_0^1\left(e^{(1+x)^2} - \frac{3e^{(1+3x)}}{2\sqrt{1+3x}}\right)dx$$

  • $\begingroup$ I am new so don't misunderstand $\endgroup$ – Mark 7 Apr 6 at 4:28
  • 3
    $\begingroup$ I suppose that in the integral it is $e^{(1\color{red}{+}x)^2}$ $\endgroup$ – Claude Leibovici Apr 6 at 5:16
  • $\begingroup$ Hint: split into two integrals and make an obvious change of variable in the second integral. $\endgroup$ – zhoraster Apr 6 at 5:21

Let $u=\sqrt{1+3x}$, then $\frac{du}{dx}=\frac{3}{2u}$

\begin{align} \frac32 \int_0^1 \frac{\exp(1+3x)}{\sqrt{1+3x}} \, dx&= \frac32 \int_1^2\frac{\exp(u^2)}{u}\cdot \frac{2u}{3}\, du = \int_1^2 \exp(u^2)\,du \end{align}

Also, let $v=1+x$,

$$\int_0^1 \exp((1+x)^2) \, dx = \int_1^2 \exp(v^2) \, dv$$

Hence, the two integral cancels out.

  • $\begingroup$ Gotcha thank you $\endgroup$ – Mark 7 Apr 6 at 5:28

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