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.

Let $a_n$, a sequence suh that $\underset{n\to \infty }{\mathop{\lim }}\,{{a}_{n}}=0$ and the series: $\sum a_n$, $\sum (a_n + a_{n+1})$

Prove/Disprove: The series converge/diverge together.

I'll be glad for an hint or a guidance.

share|improve this question
    
Thank you all for the variety of answers! it's great to see how each one of you approach the question. –  AnnieOK Apr 26 at 12:36
add comment

4 Answers 4

up vote 6 down vote accepted

$$\sum_{n=1}^N(a_n+a_{n+1})=a_{N+1}-a_1+2\sum_{n=1}^N a_n $$ and $$\sum_{n=1}^Na_n = \frac{a_0-a_{N+1}+\sum_{n=1}^N(a_1+a_{n+1})}2 $$

share|improve this answer
    
What you did is showing that each series can be written as the other multiplied by a constant. Right? That's cool! –  AnnieOK Apr 26 at 12:47
2  
As $a_{N+1}\to 0$, if we know tha the partial sums on the right converge as $N\to\infty$, then so does the partial sum on the left –  Hagen von Eitzen Apr 26 at 13:15
add comment

The second sum can be written as $$\begin{align*}\sum (a_n+a_{n+1})&=(a_1+a_2)+(a_2+a_3)+...=a_1+2a_2+2a_3+...\\\\&=-a_1+2\sum a_n \sim \sum a_n \end{align*}$$ where the symbol $\sim$ means that they behave the same as $n \to \infty$.

share|improve this answer
add comment

If $$\sum_0 ^{\infty} a_n $$does converge to some value $L$ then $$ \sum_0 ^{N} (a_n + a_{n+1}) = 2\sum_0 ^{N} a_n + a_{N+1} -a_0 \to 2L - a_0$$ As $a_n \to 0$

So conversely we know we can write $$ \sum_0 ^{N} (a_n + a_{n+1}) - a_{N+1} +a_0 = 2\sum_0 ^{N} a_n $$

And if $\sum_0 ^{N} (a_n + a_{n+1})$ converges to some $L'$ then we have $$\sum_0 ^{\infty} (a_n + a_{n+1}) - a_{N+1} +a_0 = 2\sum_0 ^{\infty} a_n = L' + a_0 $$ So then $$ \sum_0 ^{\infty} a_n = \frac {L' + a_0} {2} $$ So one of the series converges iff the other does.

share|improve this answer
2  
But $\underset{n\to \infty }{\mathop{\lim }}\,{{(-1)}^{n}}\ne 0$. Your counter-example isn't good. –  AnnieOK Apr 26 at 12:40
    
@AnnieOK sorry, missed that hypothesis, I'll edit my post –  CameronJWhitehead Apr 26 at 14:54
add comment

Hint: Consider $a_n = (-1)^n + 1/n^2$

share|improve this answer
    
How does this make $a_n\to 0$? –  Hagen von Eitzen Apr 26 at 12:30
    
@HagenvonEitzen: Ah. Didn't notice that requirement. I was showing that $\sum a_n$ could diverge while $\sum(a_n + a_{n+1})$ converges. My bad, thanks for pointing this out. –  MPW Apr 26 at 18:40
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.