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I'm asked to find out if this series converges or diverges:

$$\sum_{n=1}^{\infty} \frac{1+(-1)^n}{\sqrt{n}}=0+\frac{2}{\sqrt{2}}+0+\frac{2}{\sqrt{4}}...$$

So I thought I could use a direct comparison test, so

$$\sum_{n=1}^{\infty} \frac{1+(-1)^n}{\sqrt{n}}\leq \sum_{n=1}^{\infty}\frac{2}{\sqrt{n}}$$ But giving that this is a p-serie with $$p=-\frac{1}{2}$$ I know that I can not use this to compare with because it diverges. So I'm stuck. Does anyone have some tips?

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  • $\begingroup$ You are very close to showing divergence by a comparison test, but it should be done by a "lower bound" (actually an equality) with the even terms (equiv. with the odd terms) of the partial sums sequence. $\endgroup$ – hardmath Feb 23 at 18:21
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It diverges, since it is the sum of the divergent series $\displaystyle\sum_{n=1}^\infty\frac1{\sqrt n}$ with the convergent series $\displaystyle\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt n}$.

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  • $\begingroup$ Ohh, I have never thought about it like that. I have read though the chapter of my book, but I have not found this result. Thank you so much! $\endgroup$ – Mathomat55 Feb 23 at 18:12
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Hint:

Since $1+(-1)^n=0$ if $n$ is odd and $2$ if $n$ is even, it's

$$\sum\limits_{k=1}^\infty\dfrac{2}{\sqrt{2k}}= \sqrt2\sum\limits_{k=1}^\infty \dfrac1{\sqrt k}$$

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  • $\begingroup$ Thank you so much! It was a great explanation! $\endgroup$ – Mathomat55 Feb 23 at 18:26
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Other way to see it is that in the sum you take common factor of $\sqrt2$ u get $$\frac{1}{\sqrt2}\left(\frac{2}{\sqrt1}+\frac{2}{\sqrt2}+\frac{2}{\sqrt3}+...\right)=\frac{1}{\sqrt2}\sum_{n\geq1} \frac{2}{\sqrt n}=\frac{2}{\sqrt2}\sum_{n\geq1} \frac{1}{\sqrt n}>\infty$$

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  • $\begingroup$ Thank you so much! It was really helpful! $\endgroup$ – Mathomat55 Feb 23 at 18:27
  • $\begingroup$ Pretty sure that the last sum isn't $>\infty$... $\endgroup$ – Glen O Feb 24 at 2:23

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