# Series: prove/disprove a statement

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.

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Thank you all for the variety of answers! it's great to see how each one of you approach the question. – AnnieOK Apr 26 '14 at 12:36

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

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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 '14 at 12:47
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 '14 at 13:15

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.

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

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 $n \to \infty$.

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

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How does this make $a_n\to 0$? – Hagen von Eitzen Apr 26 '14 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 '14 at 18:40