Let $\{b_n\}$ be a sequence of positive numbers that converges to $\frac{1}{2}.$ Determine whether the given series is absolutely convergent.
$$\sum_{n=1}^{\infty}\frac{b^{n}_{n}\cos(n\pi)}{n}$$
Notice for integer $n$, $\cos(n\pi)$ alternates infinitely often between $-1$ and $1$. Thus we replace $\cos(n\pi)$ by $(-1)^n$. Our new formula becomes: $$\sum_{n=1}^{\infty}\frac{b^{n}_{n}(-1)^n}{n}$$
To test for absolute convergence, test $$\sum_{n=1}^{\infty}\left|\frac{b^{n}_{n}(-1)^n}{n}\right|=\sum_{n=1}^{\infty}\frac{b^{n}_{n}}{n}$$ for convergence.
Utilizing the Limit Comparison Test with $a_n=\frac{b^n_n}{n}$ and $b_n=\frac{1}{n}$, $$\lim_{n\to\infty}\frac{a_n}{b_n}=\lim_{n\to\infty}\frac{(b_n^n)(n)}{(n)(1)}=\frac{1}{2}>0\neq\infty$$
Since this limit exists as a positive finite value and $\sum b_n$ diverges, $\sum a_n=\sum^{\infty}_{n=1}\frac{b^n_n}{n}$ diverges.
Thus we can conclude by the definition of absolute convergence, namely a series $\sum a_n$ converges absolutely if the series $\sum\left|a_n\right|$ converges, that the series
$$\sum_{n=1}^{\infty}\frac{b^{n}_{n}\cos(n\pi)}{n}\space\text{does not absolutely converge}$$
However, multiple online sources suggest this series does converge absolutely and the answer key in my textbook suggests the same. If I am wrong, where is my error? Thanks in advance!
Best Answer
The error lies in the fact that you assumed that $\lim_{n\to\infty}{b_n}^n=\frac12$. But the hypothesis is that $\lim_{n\to\infty}b_n=\frac12$. And, yes, the series converges absolutely.