- #1
Dustinsfl
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The book says this isn't absolutely convergent but I keep getting it is by the ratio test. What is wrong?
The Fourier series for $f$ is $f(\theta) = 2\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}\sin n\theta$.
Then
$$
-\sum\limits_{n = 1}^{\infty}\left|\frac{(-1)^{n + 1}}{n}\right|.
$$
By the ratio test, we have
\begin{alignat*}{3}
\lim_{n\to\infty}\left|\frac{(-1)^{n + 2}n}{(-1)^{n + 1}(n + 1)}\right| & = & \lim_{n\to\infty}\left|\frac{-n}{n + 1}\right|\\
& = & \lim_{n\to\infty}|-1|\frac{n}{n + 1}\\
& = & 1 < \infty
\end{alignat*}
I solved this problem in another manner but shouldn't I be able to get the same answer using the Ratio Test?
The Fourier series for $f$ is $f(\theta) = 2\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n + 1}}{n}\sin n\theta$.
Then
$$
-\sum\limits_{n = 1}^{\infty}\left|\frac{(-1)^{n + 1}}{n}\right|.
$$
By the ratio test, we have
\begin{alignat*}{3}
\lim_{n\to\infty}\left|\frac{(-1)^{n + 2}n}{(-1)^{n + 1}(n + 1)}\right| & = & \lim_{n\to\infty}\left|\frac{-n}{n + 1}\right|\\
& = & \lim_{n\to\infty}|-1|\frac{n}{n + 1}\\
& = & 1 < \infty
\end{alignat*}
I solved this problem in another manner but shouldn't I be able to get the same answer using the Ratio Test?
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