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Show that $$1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{5}}-\cdots$$ is convergent but by rearrangement the following series $$\left(1+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{7}}-\frac{1}{\sqrt{4}}\right)+\cdots $$ is divergent.

Attempt:

The 1st series $$\sum_{n=1}^\infty (-1)^{n+1} \frac{1}{\sqrt{n}}$$ can be proved to be convergent by Leibnitz Test as

(1) $u_n= \frac{1}{\sqrt{n}}\to 0$ as $n\to\infty$

(2) $\{u_n\}$ is monotone decreasing

But please help me to show that the second series, after rearrangement, is divergent.

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  • $\begingroup$ I'm not sure, but can I express the term of a series in three terms?prntscr.com/jhqb92 $\endgroup$ Commented May 14, 2018 at 14:37
  • $\begingroup$ @VladislavKharlamov you're missing \ in front of your square roots $\endgroup$ Commented May 14, 2018 at 14:41

3 Answers 3

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Hint: for $n\ge 0:$ $$\dfrac{1}{\sqrt{4n+1}} + \dfrac{1}{\sqrt{4n+3}} - \dfrac{1}{\sqrt{2(n+1)}}>\dfrac{1}{\sqrt{4n+4}} + \dfrac{1}{\sqrt{4n+4}} - \dfrac{1}{\sqrt{2(n+1)}}=\\ \left(1-\frac{1}{\sqrt{2}}\right)\cdot \frac{1}{\sqrt{n+1}}$$

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Your question is: prove that the series$$\sum_{n=1}^\infty\frac1{\sqrt{4n-1}}+\frac1{\sqrt{4n-3}}-\frac1{\sqrt{2n}}$$diverges. This is true because$$\lim_{n\to\infty}\frac{\frac1{\sqrt{4n-1}}+\frac1{\sqrt{4n-3}}-\frac1{\sqrt{2n}}}{\frac1{\sqrt{n}}}=\frac12+\frac12-\frac1{\sqrt2}=1-\frac1{\sqrt2}$$and the series $\sum_{n=1}^\infty\frac1{\sqrt{n}}$ diverges.

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  • $\begingroup$ How to show that $\sum_{n=1}^\infty\frac1{\sqrt{4n-1}}+\frac1{\sqrt{4n-3}}-\frac1{\sqrt{2n}}$ is a series of positive terms, as it is essential for comparison test. $\endgroup$ Commented May 14, 2018 at 15:01
  • $\begingroup$ See my response. I made it obvious to show that the series has only positive terms. Simply set the numerator equal to zero, show that the only real root is negative, then take one positive number and show that it is positive. Since it is continuous for all positive values of $n$, it can never go from positive to negative again. $\endgroup$ Commented May 14, 2018 at 15:03
  • $\begingroup$ @user1942348 SInce the limit is greater than $0$ and the denominator is positive, the numerator is positive if $n$ is large enough, and that's enough to prove it. (Actually, every term is positive, but that's not important.) $\endgroup$ Commented May 14, 2018 at 15:04
  • $\begingroup$ @ JoséCarlosSantos I understand your point. But a mathematical proof to check the numerator positive is expected. $\endgroup$ Commented May 14, 2018 at 15:27
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    $\begingroup$ @user1942348 $$\left(\frac1{\sqrt{4n-1}}+\frac1{\sqrt{4n-1}}\right)^2>\frac1{4n-1}+\frac1{4n-3}>\frac1{4n}+\frac1{4n}=\frac1{2n}=\left(\frac1{\sqrt{2n}}\right)^2.$$ $\endgroup$ Commented May 14, 2018 at 15:31
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$$\sum_{n\ge 0} \left( \dfrac{1}{\sqrt{4n+1}} + \dfrac{1}{\sqrt{4n+3}} - \dfrac{1}{\sqrt{2(n+1)}} \right)$$

Find a common denominator:

$$\dfrac{\sqrt{(4n+3)(2n+2)}+\sqrt{(4n+1)(2n+2)}-\sqrt{(4n+1)(4n+3)}}{\sqrt{(4n+1)(4n+3)(2n+2)}}$$

Show that this is always positive (so it is not an alternating series), then use the comparison test with a divergent series to show that the series diverges. Hint: show that the fraction above is always greater than

$$\dfrac{1}{k\sqrt{n}}$$

For some arbitrarily large constant $k$.

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