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I need to classify the series $a_n=2^n-n$ as either absolutely convergent, conditionally convergent or divergent. The answer says it is divergent and says to use the nth term test to show this. I.e. Show $a_{n+1}-a_n\geq 1$ for all $n$, and hence $\lim_{n\rightarrow +\infty }a_n= +\infty $.

What I don't understand is, why have they used $a_{n+1}-a_n\geq 1$? Isn't the nth term test just testing for convergence by determining $\lim_{n\rightarrow +\infty }a_n\neq 0$? Can someone please show me how this question is done, thanks.

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  • $\begingroup$ If $a_n$ converges to $x$, then $a_{n+1}-a_n$ converges to $x-x=0$. That's all you need to understand to see why they used this. And why it implies that $|sum a_n$ diverges. $\endgroup$
    – Julien
    Apr 10, 2013 at 11:23

4 Answers 4

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A necessary condition for the convergence of a series $\displaystyle\sum_n a_n$ is $\displaystyle\lim_{n\to\infty}a_n=0$ but this is not a sufficient condition as shown in the divergent series $\displaystyle\sum_n \frac{1}{n}$, hence by contraposition if $\displaystyle\lim_{n\to\infty}a_n\not=0$ then the series $\displaystyle\sum_n a_n$ is divergent.

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  • $\begingroup$ Is there a reason it is >=1, rather than >0? $\endgroup$
    – user71865
    Apr 10, 2013 at 11:20
  • $\begingroup$ @user71865 Yes, if we suppose that the sequence $(a_n)$ has a finite limit and by passing to the limit in this inequality $a_{n+1}-a_n\geq1$ we can find a contradiction which is not the case if $1$ is replaced by $0$ $\endgroup$
    – user63181
    Apr 10, 2013 at 11:26
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The test you are referring to, says that if the series $\sum_{n=1}^\infty a_n$ converges then $a_n\xrightarrow[n\to\infty]{}0$.
Therefore if you show that $\lim_{n\to\infty}a_n\neq0$ the series $\sum_{n=1}^\infty a_n$ cannot converges.

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The answer to this question should be straight forward. Recall the property of convergence:

if $\displaystyle \sum^{\infty}_{n=0} a_n$ converges $\Rightarrow \displaystyle \lim_{n\rightarrow \infty}a_n=0$ (Remark, the converse is not true!,e.g. $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}$)

By contrapositive property in logic, it implies that:

$\displaystyle \lim_{n\rightarrow \infty}a_n \not=0 \Rightarrow \displaystyle \sum^{\infty}_{n=0} a_n$ doesn't converges, hence it diverges.

There is no other higher technique required. It's just following the basic definition of series. Hope this helps you.

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As $n \to \infty$, $a_n\to \infty$. This means that, when you are adding new terms, the suma is getting bigger and bigger. So it's divergent.

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