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Looking to prove that the following series converges conditionally

$$\sum _{n=1}^{\infty}\frac{(-1)^{n+1}(1+n)^{\frac{1}{n}}}{n}$$

Plugging in some terms I see that,

$$\sum _{n=1}^{\infty}\frac{(-1)^{n+1}(1+n)^{\frac{1}{n}}}{n} = 2 - \frac{-\sqrt{3}}{2} + \frac{^4\sqrt 4}{3}-...$$ so this series is alternating

Using the alternating series test,

let $a_n =\frac{(1+n)^{\frac{1}{n}}}{n}$

$\lim a_n = \frac{(\frac{1}{n}+ \frac{n}{n})^{\frac{1}{n}}}{\frac{n}{n}} = \lim1^{\frac{1}{n}} = 1 \ne 0$. Therefore by the Alternating series test, this diverges. If this working is correct then how do I show its conditionally convergent?

Thomas Andrews
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jh123
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  • The limit of the sequence with absolute value is zero, yet you still can not deduce the series converges conditionally. You could either show this positive sequence is monotonically descending to zero or else use Abel's Test or Dirichlet's test...or something else. – DonAntonio Feb 02 '17 at 15:36
  • @DonAntonio how can I use Abel's or Dirichlets test? – jh123 Feb 02 '17 at 15:38
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    @jh I added an answer to this. – DonAntonio Feb 02 '17 at 15:41
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    @DonAntonio thankyou for this answer, I understand now how the Abel Test works – jh123 Feb 02 '17 at 15:47

3 Answers3

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Hint: Use the squeeze theorem:

$$\frac{n^{1/n}}{n}\leq \frac{(1+n)^{1/n}}{n} \leq \frac{(2n)^{1/n}}{n}$$

MrYouMath
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Take

$$\sum_{n=1}^\infty\frac{(-1)^{n+1}}n\cdot\sqrt[n]{1+n}$$

Now, $\;\sum\limits_{n=1}^\infty\frac{(-1)^{n+1}}n\;$ converges, whereas $\;b_n:=\sqrt[n]{1+n}\;$ is monotone and bounded, thus by Abel's Test we have convergence

Added on request: It is not absolutely convergent because

$$\frac{\frac{\sqrt[n]{1+n}}n}{\frac1n}=\sqrt[n]{1+n}\xrightarrow[n\to\infty]{}1$$

and thus our series converges iff the harmonic one does (this is the limit comparison test), but the harmonic one does not converge...

DonAntonio
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  • okay thank you for your answer, we have shown that it is convergent, how do we show that it is conditionally convergent – jh123 Feb 02 '17 at 15:43
  • do I need to take the absolute value of this series now? – jh123 Feb 02 '17 at 15:44
  • @jh Yes, for absolute convergence checking you must take the absolute value of the general temr sequence. I added this to my answer... – DonAntonio Feb 02 '17 at 15:48
  • @jh Oh, dear: not at all! Will you please take a peek at https://en.wikipedia.org/wiki/Limit_comparison_test ? – DonAntonio Feb 02 '17 at 15:53
  • okay I understand what you have done here, I just got a little confused because you didn't include the limit on the equation. so taking lim an/bn gave us a positive number therefore if bn diverges which it does then an must diverge – jh123 Feb 02 '17 at 15:55
  • @jh Exactly...and I did take the limit: that $;... \xrightarrow[n\to\infty]{} 1;$ means exactly that the left side's limit is 1 when...etc. – DonAntonio Feb 02 '17 at 15:57
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    okay I see what you mean, thank you very much for your help. Abel's test makes a complicated question much easier – jh123 Feb 02 '17 at 16:00
  • and this converges to 1 because $(1+n)^{\frac{1}{n}}$ is infinity raised to the power 0 which is 1? – jh123 Feb 02 '17 at 16:14
  • @JH No, not really because of that. You can use the squeeze theorem with other well known limit: $$\sqrt[n]n\le\sqrt[n]{1+n}\le\sqrt[n]{2n}$$ and both extremes converge to $;1;$ so does our sequence. – DonAntonio Feb 02 '17 at 18:06
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By the AM-GM inequality, $$(n+1)^{1/n}=\left(\frac{2}{1}\cdot\frac{3}{2}\cdot\ldots\cdot\frac{n+1}{n}\right)^{1/n}\leq 1+\frac{H_n}{n}\leq 1+\frac{\log n}{n}\tag{1} $$ hence: $$ \sum_{n=1}^{N}\frac{(-1)^{n+1}(n+1)^{1/n}}{n} = \sum_{n=1}^{N}\frac{(-1)^{n+1}}{n} + \sum_{n=1}^{N}\frac{(-1)^{n+1}\left[(n+1)^{1/n}-1\right]}{n} \tag{2} $$ and the original series is the sum between a well-known conditionally convergent series and an absolutely convergent series (by the $p$-test). Since $\sum_{n\geq 1}\frac{(-1)^{n+1}}{n}$ is not absolutely convergent, the original series is conditionally convergent but not absolutely convergent.

Jack D'Aurizio
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