Let `< a_n >` be a sequence such that `lim_(x->oo)a_n=0.` Then `lim_(n->oo)(a_1+a_2++a_n)/(sqrt(sum_(k=1)^n k)),` is

Options

A. a.0
B. b.1
C. `c. sqrt(2)`
D. d.2

Answer

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

\[ \lim_{n \to \infty} \frac{a_1 + a_2 + \ldots + a_n}{\sqrt{\sum_{k=1}^n k}} \]

Given that \(\lim_{n \to \infty} a_n = 0\).

### Step 1: Analyze the denominator

The denominator is \(\sqrt{\sum_{k=1}^n k}\). We know that:

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

Thus, we can simplify the denominator:

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

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

\[ \sqrt{\sum_{k=1}^n k} \sim \frac{n}{\sqrt{2}} \quad \text{as } n \to \infty \]

### Step 2: Analyze the numerator

Now, consider the numerator \(a_1 + a_2 + \ldots + a_n\). Since \(\lim_{n \to \infty} a_n = 0\), it implies that the terms \(a_n\) are getting smaller. However, we need to evaluate the sum \(S_n = a_1 + a_2 + \ldots + a_n\).

Since \(a_n\) approaches 0, we can say that \(S_n\) grows slower than \(n\) as \(n\) increases. More formally, if \(a_n\) is bounded, then:

\[ S_n \leq n \cdot M \quad \text{for some constant } M \]

### Step 3: Combine the results

Now we can rewrite our limit:

\[ \lim_{n \to \infty} \frac{S_n}{\sqrt{\sum_{k=1}^n k}} = \lim_{n \to \infty} \frac{S_n}{\frac{n}{\sqrt{2}}} = \sqrt{2} \lim_{n \to \infty} \frac{S_n}{n} \]

Since \(S_n\) grows slower than \(n\) (as discussed), we can conclude that:

\[ \lim_{n \to \infty} \frac{S_n}{n} = 0 \]

### Step 4: Final result

Thus, we have:

\[ \lim_{n \to \infty} \frac{S_n}{\sqrt{\sum_{k=1}^n k}} = \sqrt{2} \cdot 0 = 0 \]

So, the final answer is:

\[ \boxed{0} \]