Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  1015 Apr 10 '13 at 11:23
add comment

4 Answers 4

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.

share|improve this answer
add comment

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.

share|improve this answer
    
Is there a reason it is >=1, rather than >0? –  user71865 Apr 10 '13 at 11:20
    
@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 $0$ –  Sami Ben Romdhane Apr 10 '13 at 11:26
add comment

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.

share|improve this answer
add comment

If a series is convergent, i.e. $\lim_{n\rightarrow \infty} < \infty$, then the terms of the series must get closer together as $n$ increases, and in fact this difference must decrease to zero. (This idea informs the notion of Cauchy convergence as well.) So, if we take a look at the difference of successive terms, i.e. $a_{n+1} - a_{n}$ and find out that the difference doesn't give us something that decreases to zero with n then we know that the series cannot converge. In your case: $$a_{n+1} - a_n = 2^{n+1} - (n+1) - (2^n - n)$$ $$ \Rightarrow a_{n+1} - a_n = 2^n - 1 $$ and this will never decrease to 0 as $n$ increases.

(A slightly easier example to see this on is $a_n = n$: then $a_{n+1} - a_n = n+1-n = 1$ and again we see that the series must diverge as 1 cannot decrease to 0).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.