# Handling a sequence in a series: $\sum_{n=1}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}$ for $a_n \rightarrow \tfrac{1}{2}$

The question is whether the series $\sum_{n=1}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}$, where $\{ a_n \}$ is a sequence of positive numbers that converges to ½, converges absolutely or not.

My hesitation is that we know nothing about the behaviour of $a_n$ (whether it is strictly increasing/decreasing or oscillatory). Here is what I am thinking, but I'm uncertain if it is sufficiently attentive to details:

$\displaystyle \sum_{n=1}^{\infty} \frac{(a_n)^n\cos(n\pi)}{n} = \sum_{n=1}^{\infty} \frac{(a_n)^n(-1)^n}{n}$

We're interested in absolute convergence, so we consider the absolute value of that series, viz:

$\displaystyle \sum_{n=1}^{\infty} \frac{(a_n)^n}{n}$

Case I: Assume $a_n \rightarrow ½$ from above, then there exists some $n > N$ such that $a_n < r$, where $r<1$, for all $n > N$. As such,

$\displaystyle \sum_{n=N}^{\infty} \frac{(a_n)^n}{n} < \sum_{n=N}^{\infty} \frac{(r)^n}{n} < \sum_{n=N}^{\infty} r^n$,

which is a convergent geometric series because $r<1$.

&nd then similarly for for $a_n \rightarrow ½$ from below. Now that I think of it, we could probably handle the entire thing in one case by using absolute values, right?

• Can't you just use Comparison Test? Sep 28 '14 at 17:16

An approach.

Let $0<\epsilon<1$. Since $\left\{a_n\right\}_{n\geq1}$ converges to $\frac 12$, then $$\exists \,N\geq0, \forall n\geq N, \left|a_n-\frac 12\right|\leq \epsilon.$$ Taking for example $\epsilon:=\frac14$, $$|a_n|\leq\left|a_n-\frac 12\right|+\frac 12\leq\frac 34, \quad n\geq N,$$ and $$|a_n|^n\leq\left(\frac{3}{4}\right)^{n}, \quad n\geq N.$$ Then you may write $$\left|\sum_{n=N}^{\infty} \frac{(a_n)^n \cos(n\pi)}{n}\right|\leq \sum_{n=N}^{\infty} \frac{1}{n}\left(\frac{3}{4}\right)^{n}$$ But on the right handside we have the rest of a convergent series, thus on the left handside the rest of our series may be as small as we want as $N$ is great.

The initial series is absolutely convergent.

• This seems to use $|a|\leqslant b+c\implies|a|^n\leqslant2b^n+2c^n$... which is wrong.
– Did
Sep 28 '14 at 17:47
• @Did Oops, fixed. Thank you! Sep 28 '14 at 18:21
• sorry for the delay in recognizing the solution - not sure what happened there. Thank you for the very clear proof! Oct 30 '15 at 22:07

We know that $\{a_n\}$ is a sequence of strictly positive numbers that converges to $\frac{1}{2}$, so we can take the absolute value of the whole thing to get rid of the cosine term, and perform the Comparison Test with the series $${\sum_{n=1}^\infty} {a_n}^n$$ which is a geometric series for all $n > N$ for some $N$ and hence converges. We know that: $${\sum_{n=1}^\infty} \frac{{a_n}^n}{n} \leq {\sum_{n=1}^\infty} {a_n}^n$$ and the result should fall out. Obviously not the most rigorous proof, and you would need to elaborate on this argument, but that should be the general gist of it.

• If $a_n\to\frac12$ then $\sum\limits_na_n$ diverges, not converges.
– Did
Sep 28 '14 at 17:48
• @ Did - Forgot to raise $a_n$ to the power of n, cheers Sep 28 '14 at 18:27
• Another problem is that $\sum\limits_n(a_n)^n$ is not a geometric series.
– Did
Sep 28 '14 at 18:31