for the power seies n^4x^n/2(n-1)...find the radius and interval of convergence
To find the radius and interval of convergence for the power series \(\sum_{n=1}^\infty \frac{n^4x^n}{2(n-1)}\), we can apply the ratio test.
The ratio test states that for a power series \(\sum_{n=1}^\infty a_n\), if the limit
\[L = \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right|\]
exists, then the series converges if \(L < 1\), diverges if \(L > 1\), and the test is inconclusive if \(L = 1\). If the ratio test is inconclusive, additional tests, such as the root test or direct comparison test, may be used.
Let's apply the ratio test to our power series:
\[a_n = \frac{n^4x^n}{2(n-1)}\]
We calculate the ratio:
\[\begin{align*}
L &= \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| \\
&= \lim_{n\to\infty} \left|\frac{\left(\frac{(n+1)^4x^{n+1}}{2n}\right)}{\left(\frac{n^4x^n}{2(n-1)}\right)}\right| \\
&= \lim_{n\to\infty} \left|\frac{(n+1)^4x^{n+1}(n-1)}{2n^4x^n}\right| \\
&= \lim_{n\to\infty} \left|\frac{(n+1)^4(n-1)}{2n^4}\right|\cdot \left|x\right|
\end{align*}\]
We can simplify this expression further:
\[\begin{align*}
L &= \lim_{n\to\infty} \left|\frac{(n^4+4n^3+6n^2+4n+1)(n-1)}{2n^4}\right|\cdot \left|x\right| \\
&= \lim_{n\to\infty} \left|\frac{n^5-5n^4+6n^3-4n^2+1}{2n^4}\right|\cdot \left|x\right| \\
&= \lim_{n\to\infty} \frac{n^5}{2n^4}\cdot \left|x\right| \\
&= \frac{1}{2}\left|x\right|
\end{align*}\]
Therefore, we have \(L = \frac{1}{2}\left|x\right|\).
To determine the interval of convergence, we know that the series converges when \(L < 1\). Thus, we solve the inequality:
\[\frac{1}{2}\left|x\right| < 1\]
By isolating \(|x|\):
\[\left|x\right| < 2\]
This means that the series converges when \(-2 < x < 2\).
Finally, to find the radius of convergence, we take the absolute value of the endpoint values:
\[\text{Radius of convergence} = \left|\frac{2 - (-2)}{2}\right| = \left|2\right| = 2\]
Therefore, the radius of convergence is 2 and the interval of convergence is \(-2 < x < 2\).