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Dtermine if the series below converges, or diverges using direct comparison test...

Question:

Dtermine if the series below converges, or diverges using direct comparison test {eq}\sum_{k=1}^{\infty} (-1)^{n} ln(1 + \frac{1}{n}){/eq}

Alternating Series Test:

Let to check for convergence and divergence of the series {eq}\sum\limits_{n = 1}^\infty {{h_n}} {/eq} .

Then series can be represnted as {eq}{h_n} = {( - 1)^{n + 1}}{q_{n\,\,\,\,\,\,\,}}{\text{or }}{h_n} = {( - 1)^n}{q_{n\,\,\,\,\,}} {/eq}, where {eq}{q_n} \geqslant 0,\forall n,\, {/eq}.

Then for convergent of the series following condition satiesfied;

(1){eq}\mathop {\lim }\limits_{n \to \infty } {q_n} = 0 {/eq}.

(2){eq}\left\{ {{q_n}} \right\} {/eq} is decreasing sequence.

Conditionally Convergent:

{eq}\displaystyle \sum {{a_k}{\text{ is conditionally convergent if }}} \sum {{a_k}{\text{ is convergent and }}} \sum {\left| {{a_k}} \right|{\text{ is divergent}}{\text{.}}} {/eq}

Answer and Explanation:

Given that: {eq}\displaystyle \sum\limits_{n = 1}^\infty {{{( - 1)}^n}} \ln \left( {1 + \frac{1}{n}} \right) {/eq}

{eq}\displaystyle \eqalign{ ...

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