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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!

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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.

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  • $\begingroup$ Oh, so the notation $b_n^n=(b_n)^n$? $\endgroup$ – coreyman317 Aug 3 '18 at 12:12
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    $\begingroup$ @coreyman317 Yes. $\endgroup$ – José Carlos Santos Aug 3 '18 at 12:13
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    $\begingroup$ $b_n^n$ is standard notation... $\endgroup$ – JavaMan Aug 3 '18 at 14:04
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You follow from $\lim\limits_{n\rightarrow\infty} b_n = \frac{1}{2}$ that $\lim\limits_{n\rightarrow\infty} b_n^n = \frac{1}{2}$, which is wrong.

But when $\lim\limits_{n\rightarrow\infty} b_n = \frac{1}{2}$, then after a certain point $N$, for all $n>N$ we have that $b_n$ is positive and $b_n < \frac{2}{3} =: \frac{1}{2}+\varepsilon$. So your series in absolute value is at most: $$\text{A finite sum} + \sum\limits_{n = 1}^{\infty}\frac{1}{(\frac{3}{2})^n\cdot n}$$

This series can be checked directly via standard methods such as the quotient test or the root test.

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We have that

$$b_n^{n}\sim \frac1{2^n}$$

therefore

$$0\le \left|\frac{b^{n}_{n}\cos(n\pi)}{n}\right|\le \frac{b^{n}_{n}}{n}\sim \frac1{n2^n}$$

and since $\sum \frac1{n2^n}$ converges also $\sum \left|\frac{b^{n}_{n}\cos(n\pi)}{n}\right|$ converges by squeeze theorem.

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