The value of `lim_(nrarroo)Sigma_(r=1)^(n)((2r)/(n^(2)))e^((r^(2))/(n^(2)))` is equal to

Options

A. e
B. 2e
C. `e-2`
D. `e-1`

Answer

To solve the limit

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2r}{n^2} e^{\frac{r^2}{n^2}}, \]

we can convert the summation into an integral. Here’s the step-by-step solution:

### Step 1: Rewrite the Summation
We start with the expression:

\[ \sum_{r=1}^{n} \frac{2r}{n^2} e^{\frac{r^2}{n^2}}. \]

We can factor out \(\frac{1}{n^2}\):

\[ \frac{1}{n^2} \sum_{r=1}^{n} 2r e^{\frac{r^2}{n^2}}. \]

### Step 2: Change of Variable
Next, we introduce a change of variable. Let \(x = \frac{r}{n}\). Then, as \(r\) goes from \(1\) to \(n\), \(x\) goes from \(\frac{1}{n}\) to \(1\). The increment \(dr\) can be expressed as \(n \, dx\).

Thus, we rewrite the summation:

\[ \sum_{r=1}^{n} 2r e^{\frac{r^2}{n^2}} = \sum_{r=1}^{n} 2(n x) e^{(n^2 x^2)/n^2} = 2n \sum_{r=1}^{n} x e^{x^2}. \]

### Step 3: Convert to Integral
Now, we can express the sum as an integral in the limit as \(n \to \infty\):

\[ \lim_{n \to \infty} \frac{1}{n^2} \cdot 2n \sum_{r=1}^{n} x e^{x^2} \cdot n \, dx \approx \int_0^1 2x e^{x^2} \, dx. \]

### Step 4: Evaluate the Integral
Now we need to evaluate the integral:

\[ \int_0^1 2x e^{x^2} \, dx. \]

Using the substitution \(u = x^2\), we have \(du = 2x \, dx\). Changing the limits accordingly, when \(x = 0\), \(u = 0\) and when \(x = 1\), \(u = 1\):

\[ \int_0^1 e^{x^2} \, dx = \int_0^1 e^u \, du = e^1 - e^0 = e - 1. \]

### Final Result
Thus, the value of the limit is:

\[ \lim_{n \to \infty} \sum_{r=1}^{n} \frac{2r}{n^2} e^{\frac{r^2}{n^2}} = e - 1. \]