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Does the series $\sum_{n\ge1}\left(-1\right)^n\frac{\cos\left(\alpha n\right)} {\sqrt n}$ converge ?

I tried to deal with this problem this way. Let $S_k$ be a sequence of partial sums of the given series.

Than $S_{2k}=\sum_{k=2}^{2n} \frac{\cos\left(\alpha n\right)} {\sqrt n}$, series $\sum_{k=2}^\infty \frac{\cos\left(\alpha n\right)} {\sqrt n}$ converges by Dirichlet Convergence Test, therefore $S_{2k}$ converges.

And $S_{2k+1}=-\sum_{k=1}^{2n+1} \frac{\cos\left(\alpha n\right)} {\sqrt n}$.

If $S_{2k}\to 0$ than also $S_{2k+1}\to 0$ and the given series converges, but I do not think that $S_{2k}$ must converge to zero.

Are there another approach to this problem ?


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$(-1)^n\cos(\alpha n) = \cos((\alpha+\pi)n)$. – achille hui Jan 9 '14 at 9:51
If $\alpha = \pi$ the series diverges. – TonyK Jan 9 '14 at 10:54
up vote 4 down vote accepted

Answer. The series converges iff $a\ne (2k+1)\pi$.

This can be proved using Abel's summation method, since it is a series of the form $$ \sum_{n=1}^\infty a_nb_n $$ with $a_n=\frac{1}{\sqrt{n}}$ decreasing and tending to zero, and $b_n=(-1)^n\cos(an)$ which has bounded partial sums.

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How can we show that $b_n$ is bounded? :s – Marko Karbevski Jul 31 '14 at 23:58
@MarkoKarbevski: $$\lvert b_n\rvert=\lvert\cos(an)\rvert\le 1.$$ – Yiorgos S. Smyrlis Aug 1 '14 at 15:58
But that does not make $b_n$ to have bounded partial sums i.e. it does not make the partial sums $\sum_{n=1}^{k} b_n$ bounded for all $k \in \mathbb{N}$. – Marko Karbevski Aug 1 '14 at 16:47

We know that $\cos(\alpha n$) oscillates between $-1$ and $1$. So, for now, we look at the following series,

$$\sum_{n=1}^{\infty} (-1)^n \frac{a}{\sqrt{n}}$$

The series shown above converges because of the alternating series test. You must prove:

$$(1) \lim \limits_{x \to \infty} \frac{a}{\sqrt{n}}=0$$ $$(2)\frac{a}{\sqrt{n}} \text{is decreasing}$$ $$(3) \frac{a}{\sqrt{n}}>0$$ All three are obviously true in this case. Hence, the series is convergent. Notice that its conditionally convergent since,

$$\sum_{n=1}^{\infty} \frac{a}{\sqrt{n}} \text{is divergent}$$

So, now we come back to $\cos(\alpha n)$. The function $\cos(\alpha n)$ if $\alpha=(2b+1)\pi$ for $n,b \in \mathbb{Z}$ and $n>1$ is equivalent to $(-1)^n$. Hence, if if $\alpha=(2b+1)\pi$ when $n,b \in \mathbb{Z}$ then the series is divergent: $$\sum_{n=1}^{\infty} (-1)^n \frac{(-1)^n}{\sqrt{n}}=\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$$

The next question to answer is what if $\alpha \neq (2b+1)\pi$

It would most definitely converge when $\alpha=0$

This series is the real part of $\sum\limits_{n\geqslant1}n^{-a}z^n$. Dirichlet's test would be appropriate to prove that it is convergent.

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