absolute convergent, conditionally convergent or divergent?

Question

I have to find out if $\displaystyle\sum_{n=2}^{\infty}$$\dfrac{\cos(\frac{\pi n}{2}) }{\sqrt n \log(n) }$ is absolute convergent, conditional convergent or divergent. I think it's divergent while the value for $\cos\left(\dfrac{\pi n}{2}\right)$ swings between $0$, $1$ and $-1$. And for $\left|\cos\left(\dfrac{\pi n}{2}\right)\right|$ it still swings between $0$ and $1$. But how can I show it formally?

Answer 1

Your series is$$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{2n}\log(2n)},$$which converges by the Leibniz test.

Answer 2

Hint:

$$\cos\frac{\pi n}2=\begin{cases}&0,&n\text{ is odd}\\{}\\ &\!\!\!\!\!-1,&n=2\pmod 4\\{}\\&1,&n=0\pmod4\end{cases}$$

So we can write

$$\sum_{n=1}^\infty\frac{\cos\frac{\pi n}2}{\sqrt n\,\log n}=\sum_{n=1}^\infty\frac{\cos\frac{2\pi n}2}{\sqrt{2n}\,\log2n}=\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{2n}\log2n}$$

Now it is easy to see the series is convergent as it is a Leibniz series when we express it as above, and with absolute value the series general term is

$$\frac1{\sqrt{2n}\log2n}\ge\frac1{\sqrt{2n}\sqrt{2n}}=\frac1{2n}$$

and the comparison test gives divergence. End the argument now.

Answer 3

By Dirichlet's test, take $a_k = \frac{1}{\sqrt{k} \log k}$ is monotone and converges to 0, $\sum_{k=1}^{n} b_k = \sum_{k=1}^{n} \cos \frac{\pi k }{2} <1=M \ \forall n$, so $\sum_{k=1}^{\infty} \cos \frac{\pi k}{2} \frac{1}{\sqrt{k} \log k}$ converges.

Answer 4

For $N \geq 2$, let $$ S_N = \sum_{n=2}^N \frac{ \cos( n \cdot \pi/2)}{\sqrt{n} \, \ln n} $$ be the partial sum up to index $n$. The given series is said to converge of the sequence $(S_n)_n$ converges.

Observe that $\cos(n \cdot \pi/2)$ is zero for odd $n$, so $S_{2n} = S_{2n+1}$ for all $n \geq 1$. So in fact, we may restrict attention to the subsequence $(S_{2n})_n$. Equivalently, we consider the partial sums $$ T_N = \sum_{n=2}^{N} \frac{ \cos(n \cdot \pi)}{\sqrt{2n} \,\ln(2n)} \text{.} $$

Since $$ \cos(n \cdot \pi) = \begin{cases} 1 &, n \text{ even}, \\ -1 &, n \text{ odd} \end{cases} \text{,} $$ this could be an alternating series. To this, we verify $\frac{1}{\sqrt{2n} \ln(2n)} > 0$. The square root function is positive unless its input is zero, which does not happen because $0$ is not a value taken by the index. $\ln(2n) > 0$ as long as $n > 1/2$, which is true for every value of the index. Therefore, the terms of this series alternate in sign.

To apply the alternating series test, we must verify that the magnitudes of the terms are (eventually) monotonically decreasing and have limit zero. First, $$ \frac{\mathrm{d}}{\mathrm{d}n} \frac{1}{\sqrt{2n} \,\ln(2n)} = \frac{ -2 -\ln(2n)}{2\sqrt{2} n \sqrt{n} \ln^2(2n)} \text{.} $$ As observed above, the index is always positive and greater than $1/2$, so the numerator is negative and the denominator is positive. This function is (strictly) monotonically decreasing on $[2,\infty)$, so the sequence of magnitudes of terms is also decreasing. It is immediate that $$ \lim_{n \rightarrow \infty} \frac{1}{\sqrt{2n} \,\ln(2n)} = 0 \text{.} $$

Its hypotheses having been met, we apply the alternating series test and conclude the series whose partial sums are $(T_n)_n$ converges. Therefore, the given series (whose partial sums are $(S_n)_n$, converges. We have concluded that the series is either absolutely or conditionally convergent, as the sequence of partial sums $$ U_N = \sum_{n=2}^{N} \frac{1}{\sqrt{2n} \,\ln(2n)} $$ converges or diverges.

At $n = 2$, $\sqrt{2n} \ln(2n) = 2\ln 4 < 4 = 2n$. Then, for $n \geq 2$, $\frac{\mathrm{d}}{\mathrm{d}n} \sqrt{2n} \ln(2n) = \frac{2 + \ln(2n)}{\sqrt{2n}}$ which has maximum $1 + \ln(2)$ at $n = 2$ and is always less than $\frac{\mathrm{d}}{\mathrm{d}n} 2n = 2$. Therefore, $\sqrt{2n} \ln(2n) < 2n$, and $$ \frac{1}{\sqrt{2n} \ln(2n)} > \frac{1}{2n} $$ for $n \geq 2$. By direct comparison, the series whose partial sums are $(U_n)_n$ diverges if $\sum_{n=2}^\infty \frac{1}{2n}$ diverges, which it does. (Its sequence of partial sums is the seuqence of halves of the partial sums of the diverging $p$-series $\sum \frac{1}{n}$. Since the $p$-series partial sums diverge, so do their halves.)

Therefore, the give series conditionally converges.