```
Determine whether the sequence converges or diverges. If it converges, find the DNE.)
$\begin{array}{l}
a_{n}=\frac{n}{5} \sin (5 / n) \\
\lim _{n \rightarrow \infty} a_{n}=\square
\end{array}$
```

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Answer from Sia

Posted 25 days ago

Solution by Steps

step 1

We start by analyzing the given sequence $a_n = \frac{n}{5} \sin \left( \frac{5}{n} \right)$

step 2

To find the limit as $n$ approaches infinity, we use the fact that $\sin(x) \approx x$ when $x$ is close to 0. Here, as $n \to \infty$, $\frac{5}{n} \to 0$

step 3

Substitute $\sin \left( \frac{5}{n} \right) \approx \frac{5}{n}$ into the sequence: $a_n \approx \frac{n}{5} \cdot \frac{5}{n}$

step 4

Simplify the expression: $a_n \approx \frac{n}{5} \cdot \frac{5}{n} = 1$

step 5

Therefore, $\lim_{n \to \infty} a_n = 1$

Answer

The sequence converges to 1.

Key Concept

Limit of a sequence involving trigonometric functions

Explanation

By approximating $\sin(x)$ with $x$ for small values of $x$, we can simplify the sequence and find its limit as $n$ approaches infinity.

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