Question

If \(\sum {{a_n}} \)is a convergent series with positive terms, is it true\(\sum {{\mathop{\rm Sin}\nolimits} \left( {{a_n}} \right)} \)is also convergent.

Short answer

To prove that\(\sum {{\mathop{\rm Sin}\nolimits} \left( {{a_n}} \right)} \)is convergent. We use limit comparison test for\(\sum {{\mathop{\rm Sin}\nolimits} \left( {{a_n}} \right)} \).

Step-by-step solution

Given data:

Since\(\sum {{a_n}} \)is convergent.

Then\(\mathop {\lim }\limits_{n \to \infty } {a_n} = 0\).

Now, we use limit comparison test for\(\sum {{\mathop{\rm Sin}\nolimits} \left( {{a_n}} \right)} \).

We find\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {{a_n}} \right)}}{{{a_n}}}\).

Finding\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {{a_n}} \right)}}{{{a_n}}}\):

For an instant, we assume that\(x = {a_n}\).

Soas\(n \to \infty \).

Then\(\mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {{a_n}} \right)}}{{{a_n}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\sin \left( x \right)}}{x} = 1\left( {\mathop {\lim }\limits_{\theta \to 0} \frac{{{\mathop{\rm Sin}\nolimits} \theta }}{\theta } = 1} \right)\)

\( \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \left( {{a_n}} \right)}}{{{a_n}}} = 1 > 0\).

Since\(\sum {{a_n}} \)is convergent so\(\sum {{\mathop{\rm Sin}\nolimits} \left( {{a_n}} \right)} \)must be convergent.

Yes, it is true,\(\sum {{\mathop{\rm Sin}\nolimits} \left( {{a_n}} \right)} \)is a convergent series.

Thus\(\sum {{\mathop{\rm Sin}\nolimits} \left( {{a_n}} \right)} \)is convergent.