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I would like to prove using Taylor expansion that the series $\sum\left(\sqrt{1+\frac{(-1)^n}{\sqrt{n}}}-1\right)$ is divergent for $n\geq 1$. What is the expansion to prove it ?

Thanks

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  • $\begingroup$ What makes you think you can prove this with a Taylor Series $\endgroup$ – Stella Biderman Feb 14 '16 at 15:42
  • $\begingroup$ I'm working with a math book which explains to do it this way without giving the answer $\endgroup$ – tomas Feb 14 '16 at 15:43
  • $\begingroup$ This is a nice problem is the sense that it' s easy to fool oneself that the series converges if we for example write it on the form $\frac {(-1)^n} {\sqrt {n}} \frac {1} {\sqrt {1 + \frac {(-1)^n} {\sqrt {n}}} + 1} $. The last factor is $\simeq \frac {1} {2} $ so one could naively think that the series should be close to $\frac {(-1)^n} {2\sqrt {n}} $ which converges. However the alternating series test does not apply to the series in question as $\frac {1} {\sqrt {n}} \frac {1} {\sqrt {1 + \frac {(-1)^n} {\sqrt {n}}} + 1}$ does not decrease monotonely. $\endgroup$ – Winther Feb 15 '16 at 18:05
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Hint. You may use, as $x \to 0$, the Taylor expansion $$ \sqrt{1+x}-1=\frac{x}{2}-\frac{x^2}{8}+O(x^3) $$ giving, for some great $n_0$, and for any $N$ greater than $n_0$,

$$ \sum_{n_0 \leq n \leq N } \left( \sqrt{1+\frac{(-1)^n}{\sqrt{n}}}-1\right)=\sum_{n_0 \leq n \leq N } \frac{(-1)^n}{2\sqrt{n}}-\sum_{n_0 \leq n \leq N } \frac{1}{8n}+\sum_{n_0 \leq n \leq N } O\left( \frac{1}{n^{3/2}}\right) $$

then letting $N \to \infty$ leads to the divergence of the series on the left hand side.

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  • $\begingroup$ J'adore toujours vos réponses, vous le faites avec tellement d'élégance, c'est fabuleux ! $\endgroup$ – Hexacoordinate-C Feb 15 '16 at 17:59
  • $\begingroup$ @Shadock Merci pour ce compliment, qui me touche. $\endgroup$ – Olivier Oloa Feb 15 '16 at 20:03
  • $\begingroup$ Il n'y a pas de quoi, quand je vois la solution de Marthy j'ai envie de dire mais oui bien-sûr haha. France 1 vs EU 0 :-) $\endgroup$ – Hexacoordinate-C Feb 15 '16 at 23:11
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$\begin{array}\\ \sum_{n=1}^{2m} \ \sqrt{1+\frac{(-1)^n}{\sqrt{n}}}-1 &=\sum_{n=1}^{2m} \left( \sqrt{1+\frac{(-1)^n}{\sqrt{n}}}-1 \right) \frac{\sqrt{1+\frac{(-1)^n}{\sqrt{n}}}+1}{\sqrt{1+\frac{(-1)^n}{\sqrt{n}}}+1}\\ &=\sum_{n=1}^{2m} \frac{\frac{(-1)^n}{\sqrt{n}}}{\sqrt{1+\frac{(-1)^n}{\sqrt{n}}}+1}\\ &=\sum_{n=1}^{2m} \frac{(-1)^n}{\sqrt{n}\sqrt{1+\frac{(-1)^n}{\sqrt{n}}}+1}\\ &=\sum_{n=1}^{2m} \frac{(-1)^n}{\sqrt{n+(-1)^n\sqrt{n}}+1}\\ &=\sum_{n=1}^{m} \frac{(-1)^{2n}}{\sqrt{2n+(-1)^{2n}\sqrt{2n}}+1} +\sum_{n=1}^{m} \frac{(-1)^{2n-1}}{\sqrt{2n-1+(-1)^{2n-1}\sqrt{2n-1}}+1}\\ &=\sum_{n=1}^{m} \frac{1}{\sqrt{2n+\sqrt{2n}}+1} -\sum_{n=1}^{m} \frac{1}{\sqrt{2n-1-\sqrt{2n-1}}+1}\\ &=\sum_{n=1}^{m}\left( \frac{1}{\sqrt{2n+\sqrt{2n}}+1} - \frac{1}{\sqrt{2n-1-\sqrt{2n-1}}+1}\right)\\ &=\sum_{n=1}^{m} \frac{(\sqrt{2n-1-\sqrt{2n-1}}+1)-(\sqrt{2n+\sqrt{2n}}+1)}{(\sqrt{2n+\sqrt{2n}}+1)(\sqrt{2n-1-\sqrt{2n-1}}+1)}\\ &=\sum_{n=1}^{m} \frac{\sqrt{2n-1-\sqrt{2n-1}}-\sqrt{2n+\sqrt{2n}}}{(\sqrt{2n+\sqrt{2n}}+1)(\sqrt{2n-1-\sqrt{2n-1}}+1)}\\ \end{array} $

The denominator is $2n+O(\sqrt{n})$.

The numerator is

$\begin{array}\\ \sqrt{2n-1-\sqrt{2n-1}}-\sqrt{2n+\sqrt{2n}} &=\sqrt{2n}\left(\sqrt{1-1/(2n)-\sqrt{1/(2n)-1/(2n)^2}}-\sqrt{1+1/\sqrt{2n}}\right)\\ &=\sqrt{2n}\left(\sqrt{1-1/(2n)-\sqrt{1/(2n)-1/(2n)^2}} -(1+1/(2\sqrt{2n})+O(1/n))\right)\\ &=\sqrt{2n}\left(\sqrt{1-\sqrt{1/(2n)}(\sqrt{1-1/(2n)}+1/\sqrt{2n}} -(1+1/(2\sqrt{2n})+O(1/n))\right)\\ &=\sqrt{2n}\left(\sqrt{1-\sqrt{1/(2n)}((1-1/(4n)+O(1/n^2))+1/\sqrt{2n}} -(1+1/(2\sqrt{2n})+O(1/n))\right)\\ &=\sqrt{2n}\left(\sqrt{1-\sqrt{1/(2n)}((1+1/\sqrt{2n}+O(1/n)} -(1+1/(2\sqrt{2n})+O(1/n)))\right)\\ &=\sqrt{2n}\left((1-\sqrt{1/(8n)}((1+1/\sqrt{2n}+O(1/n)) -(1+1/(2\sqrt{2n})+O(1/n)))\right)\\ &=-\sqrt{2n}\left(2\sqrt{1/(8n)}+O(1/n)\right)\\ &=-1+O(1/\sqrt{n})\\ \end{array} $

Therefore each term is $\frac{-1+O(1/\sqrt{n})}{2n+O(\sqrt{n})} =-\frac1{2n}+O(n^{-3/2}) $ and the sum of these diverges.

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    $\begingroup$ I found an error $\endgroup$ – Hexacoordinate-C Feb 15 '16 at 23:13
  • $\begingroup$ So. What was it? $\endgroup$ – marty cohen Feb 15 '16 at 23:47
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    $\begingroup$ It was a joke, I will not try to verify, it is a bit hard to read, you should erase the align mod in your second step because it creates "interferences" with the links at the right of the page :) $\endgroup$ – Hexacoordinate-C Feb 15 '16 at 23:51

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