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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?

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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. – 2017-02-02
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    @DonAntonio how can I use Abel's or Dirichlets test? – 2017-02-02
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    @jh I added an answer to this. – 2017-02-02
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    @DonAntonio thankyou for this answer, I understand now how the Abel Test works – 2017-02-02

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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.

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Hint: Use the squeeze theorem:

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

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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...

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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 – 2017-02-02
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    do I need to take the absolute value of this series now? – 2017-02-02
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    @jh Yes, for absolute convergence checking you must take the absolute value of the general temr sequence. I added this to my answer... – 2017-02-02
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    @jh Oh, dear: not at all! Will you please take a peek at https://en.wikipedia.org/wiki/Limit_comparison_test ? – 2017-02-02
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    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 – 2017-02-02
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    @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. – 2017-02-02
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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 – 2017-02-02
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    and this converges to 1 because $(1+n)^{\frac{1}{n}}$ is infinity raised to the power 0 which is 1? – 2017-02-02
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    @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. – 2017-02-02