Show that `lim_(nrarr infty) sum_(k=0)^(n)(""^(n)Ck)/(n^k(k+3))=e-2`

Answer

To solve the limit

\[ \lim_{n \to \infty} \sum_{k=0}^{n} \frac{{n \choose k}}{n^k (k+3)} = e - 2, \]

we will follow a series of steps to manipulate the expression and evaluate the limit.

### Step 1: Rewrite the summation
We start by rewriting the summation in a more manageable form. We can express the term \(\frac{1}{k+3}\) as an integral:

\[ \frac{1}{k+3} = \int_0^1 x^{k+2} \, dx. \]

Thus, we can rewrite the original limit as:

\[ \lim_{n \to \infty} \sum_{k=0}^{n} \frac{{n \choose k}}{n^k} \int_0^1 x^{k+2} \, dx. \]

### Step 2: Interchange summation and integration
We can interchange the summation and the integral (justified by Fubini's theorem):

\[ \lim_{n \to \infty} \int_0^1 x^{2} \sum_{k=0}^{n} \frac{{n \choose k}}{n^k} x^k \, dx. \]

### Step 3: Simplify the summation
The summation \(\sum_{k=0}^{n} {n \choose k} \left(\frac{x}{n}\right)^k\) can be recognized as the binomial expansion:

\[ \sum_{k=0}^{n} {n \choose k} \left(\frac{x}{n}\right)^k = \left(1 + \frac{x}{n}\right)^n. \]

Thus, we can rewrite our limit as:

\[ \lim_{n \to \infty} \int_0^1 x^{2} \left(1 + \frac{x}{n}\right)^n \, dx. \]

### Step 4: Evaluate the limit
As \(n \to \infty\), \(\left(1 + \frac{x}{n}\right)^n\) approaches \(e^x\). Therefore, we have:

\[ \lim_{n \to \infty} \int_0^1 x^{2} e^x \, dx. \]

### Step 5: Calculate the integral
Now we need to compute the integral:

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

We can use integration by parts. Let \(u = x^2\) and \(dv = e^x dx\). Then \(du = 2x dx\) and \(v = e^x\). Applying integration by parts:

\[ \int u \, dv = uv - \int v \, du, \]

we get:

\[ \int_0^1 x^2 e^x \, dx = \left[ x^2 e^x \right]_0^1 - \int_0^1 e^x (2x) \, dx. \]

Evaluating the boundary term:

\[ \left[ x^2 e^x \right]_0^1 = 1^2 e^1 - 0 = e. \]

Now we need to compute \(\int_0^1 2x e^x \, dx\) using integration by parts again. Let \(u = 2x\) and \(dv = e^x dx\). Then \(du = 2 dx\) and \(v = e^x\):

\[ \int 2x e^x \, dx = 2 \left[ x e^x \right]_0^1 - \int_0^1 2 e^x \, dx. \]

Evaluating the boundary term:

\[ 2 \left[ x e^x \right]_0^1 = 2(e - 0) = 2e. \]

Now, we compute \(\int_0^1 2 e^x \, dx\):

\[ \int_0^1 2 e^x \, dx = 2 \left[ e^x \right]_0^1 = 2(e - 1). \]

Putting everything together:

\[ \int_0^1 2x e^x \, dx = 2e - 2(e - 1) = 2e - 2e + 2 = 2. \]

Thus,

\[ \int_0^1 x^2 e^x \, dx = e - 2. \]

### Final Result
Putting it all together, we find:

\[ \lim_{n \to \infty} \sum_{k=0}^{n} \frac{{n \choose k}}{n^k (k+3)} = e - 2. \]

Hence, we have shown that

\[ \lim_{n \to \infty} \sum_{k=0}^{n} \frac{{n \choose k}}{n^k (k+3)} = e - 2. \]