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Does the following series converge?

$$\sum_{n=1}^{\infty}\frac{\cos\left({\frac{n\pi}{2}}\right)}{\sqrt{n}}$$

The $\cos$ function:

  • alternates between (-1) and 1 for every $n$ that is even. (for a general expression see this)
  • equal to zero for every $n$ that is odd.

Thus the sequence of the given series is actually alternating between positive and negative:

$$ \sum_{n=1}^{\infty}\frac{\cos\left({\frac{n\pi}{2}}\right)}{\sqrt{n}} = 0 + \frac{-1}{\sqrt{2}} + 0 + \frac{1}{\sqrt{4}} + 0 + \frac{-1}{\sqrt{6}} + \cdots + \frac{i^{n} (1 + (-1)^{n})}{2}\cdot\frac{1}{\sqrt{n}} $$

Would it be valid to use the Leibniz Test in order to say that the given series converge?

Do you know of a different method to prove whether the given series converges/diverges? Without using the complex number $i$.

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    $\begingroup$ The series converges, as you argue, but not to $0$. $\endgroup$ – David Mitra Dec 12 '14 at 9:30
  • $\begingroup$ You might want to take a look at Dirichlet eta function, which provides the summation of this series en.wikipedia.org/wiki/Dirichlet_eta_function $\endgroup$ – Peter Kravchuk Dec 12 '14 at 10:03
  • $\begingroup$ The sum is equal to $\left(1-\frac{1}{\sqrt{2}}\right) \zeta \left(\frac{1}{2}\right) \simeq -0.427728$ $\endgroup$ – Dr. Wolfgang Hintze Oct 14 '18 at 18:35
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Dor, you can write the following $$\displaystyle a_n=\cos\left (\frac{\pi n}{2}\right)=\begin{cases} 0, & \mathrm{n \ odd} \\ (-1)^{\large \frac{n}{2}}, & \mathrm{n \ even}\end{cases}$$ Hence, $\displaystyle a_{2k-1}=0 \ , \ a_{2k}=(-1)^k$.

Assume that the series does converge, then we can write $$\displaystyle \sum_{n=1}^{\infty} a_n=\underbrace{\sum_{k=1}^{\infty}a_{2k+1}\cdot{\frac{1}{\sqrt{2k+1}}}}_{=0}+\sum_{k=1}^{\infty}a_{2k}\cdot{\frac{1}{\sqrt{2k}}}=0+\sum_{k=1}^{\infty}(-1)^k\cdot{\frac{1}{\sqrt{2k}}}$$

Look at the sequence $a_k=\frac{1}{\sqrt{2k}}$. It is monotonically decreasing to $0$, thus, by Leibniz criterion we can write that $\displaystyle \sum_{k=1}^{\infty}(-1)^k\cdot{\frac{1}{\sqrt{2k}}}$ converges, therfore $\displaystyle \sum_{n=1}^{\infty}\frac{\cos\left(\frac{\pi n}{2}\right)}{\sqrt{n}}$ converges.

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In the Leibniz Test, for the convergence of $\sum(-1)^n a_n$ is required $a_n\ge 0$, $a_n\to 0$ and $a_{n+1}\le a_n$, so take only the even terms. And the sum isn't zero.

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$\frac{1}{\sqrt n}\downarrow\to0$and we have a constant $M$make$$ \left|\sum_{k=1}^{n}\cos\frac{k\pi}{2}\right|\leq M $$ so according to Dirchlet Thm,the series is converge.

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Is your series maybe equal to

$\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{2n}}$ (as you pointed out indirectly)?

If so, note that the nth term of the new series is decreasing and alternating as n increases.

Edit: From where does the i come?

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You can look at the partial sums $s_{2m} = \sum_{k=1}^{2m}\cos(k\pi/2)/\!\sqrt{k}$ and $s_{2m+1}$. Use the equality $$ \sum_{k=1}^{2m} a_k = \sum_{k=1}^{m}a_{2k} + \sum_{k=1}^{m} a_{2k-1}. $$ Prove that $s_{2m}$ converges and $\lim\,(s_{2m+1}-s_{2m})=0$. This is enough to show that $\lim s_{2m+1}$ exists and equals $\lim s_{2m}$. Although you cannot use $\lim(a_n \pm b_n) = \lim a_n \pm \lim b_n$, since we don't know if $\lim s_{2m+1}$ exists, you can add $\lim s_{2m}$ to both sides of $\lim\,(s_{2m+1}-s_{2m})=0$ to get $\lim s_{2m}=\lim s_{2m+1}$. Therefore, $s_{m}$ converges (why?).

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