If `S_n=sum_(k=1)^n a_k and lim_(n->oo)a_n=a ,` then `lim_(n->oo)(S_(n+1)-S_n)/sqrt(sum_(k=1)^n k)` is equal to

Options

A. 0
B. a
C. `sqrt(2)a`
D. 2a

Answer

To solve the problem, we need to evaluate the limit:

\[ \lim_{n \to \infty} \frac{S_{n+1} - S_n}{\sqrt{\sum_{k=1}^n k}} \]

where \( S_n = \sum_{k=1}^n a_k \) and \( \lim_{n \to \infty} a_n = a \).

### Step 1: Express \( S_{n+1} - S_n \)

We know that:

\[ S_{n+1} = S_n + a_{n+1} \]

Thus, we can write:

\[ S_{n+1} - S_n = a_{n+1} \]

### Step 2: Write the limit expression

Substituting \( S_{n+1} - S_n \) into the limit, we have:

\[ \lim_{n \to \infty} \frac{S_{n+1} - S_n}{\sqrt{\sum_{k=1}^n k}} = \lim_{n \to \infty} \frac{a_{n+1}}{\sqrt{\sum_{k=1}^n k}} \]

### Step 3: Evaluate \( \sum_{k=1}^n k \)

The sum of the first \( n \) natural numbers is given by:

\[ \sum_{k=1}^n k = \frac{n(n+1)}{2} \]

Thus, we can rewrite the limit as:

\[ \lim_{n \to \infty} \frac{a_{n+1}}{\sqrt{\frac{n(n+1)}{2}}} \]

### Step 4: Simplify the denominator

The square root of the sum becomes:

\[ \sqrt{\frac{n(n+1)}{2}} = \sqrt{\frac{n^2 + n}{2}} = \sqrt{\frac{n^2(1 + \frac{1}{n})}{2}} = \frac{n}{\sqrt{2}} \sqrt{1 + \frac{1}{n}} \]

As \( n \to \infty \), \( \sqrt{1 + \frac{1}{n}} \to 1 \). Therefore, we have:

\[ \sqrt{\frac{n(n+1)}{2}} \sim \frac{n}{\sqrt{2}} \text{ as } n \to \infty \]

### Step 5: Substitute back into the limit

Now substituting this back into our limit expression, we get:

\[ \lim_{n \to \infty} \frac{a_{n+1}}{\frac{n}{\sqrt{2}}} = \lim_{n \to \infty} \frac{\sqrt{2} a_{n+1}}{n} \]

### Step 6: Evaluate the limit

Since \( \lim_{n \to \infty} a_n = a \), we have \( \lim_{n \to \infty} a_{n+1} = a \). Thus, we can write:

\[ \lim_{n \to \infty} \frac{\sqrt{2} a_{n+1}}{n} = \lim_{n \to \infty} \frac{\sqrt{2} a}{n} = 0 \]

### Conclusion

Therefore, the final result is:

\[ \lim_{n \to \infty} \frac{S_{n+1} - S_n}{\sqrt{\sum_{k=1}^n k}} = 0 \]