is the series from 1 to infinity of ( 6x)/(sqrt(25x^4-1)) convergent? mention tests used

To determine the convergence of the series 

\[
\sum_{n=1}^{\infty} \frac{6x}{\sqrt{25x^4 - 1}}
\]

we need to examine the behavior of the general term:

\[
a_n = \frac{6x}{\sqrt{25x^4 - 1}}.
\]

Given that \(x\) does not vary with \(n\), the term \(a_n\) is a constant and does not depend on \(n\). Therefore, it cannot converge to zero as \(n \to \infty\) (unless \(x = 0\), in which case the whole series would trivially be zero). 

### 1. Divergence Test (or nth-term Test for Divergence)

For a series \(\sum a_n\) to converge, a necessary condition is that \(\lim_{n \to \infty} a_n = 0\). Since \(a_n = \frac{6x}{\sqrt{25x^4 - 1}}\) is constant (for any nonzero \(x\)), it does not approach zero as \(n \to \infty\).

Hence, by the **Divergence Test**, the series diverges for all \(x \neq 0\).

### Conclusion
Since the Divergence Test shows that the terms do not approach zero, the series \(\sum_{n=1}^{\infty} \frac{6x}{\sqrt{25x^4 - 1}}\) **diverges** for any nonzero \(x\).

Would you like more details or have any questions?

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1. What is the Divergence Test and why is it used in series analysis?
2. How do we apply other convergence tests, like the Ratio Test, to more complex series?
3. What happens if \(a_n\) is not constant but depends on \(n\)?
4. How would a term that does approach zero affect the series' convergence?
5. Can other tests be applied to this specific series to further verify divergence?

**Tip:** The Divergence Test is a quick first step in testing for convergence—it helps determine if further testing is even necessary!

Is the series from 1 to infinity of (6x)/(sqrt(25x^4 - 1)) convergent? Mention tests used.

Analyze the convergence of the infinite series (6x)/(sqrt(25x^4 - 1)) by applying the Divergence Test. Explore series convergence principles, including when the nth term does not approach zero. Suitable for college-level calculus students.

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