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A sequence $a_n$ is given such that $a_n \rightarrow 0\wedge a_n>0\ \forall n$.
I am to proof that
$$\left|\sum_{n=1}^{\infty}(-1)^{n+1}a_n - \sum_{n=1}^{N}(-1)^{n+1}a_n\right| \le a_{N+1}.$$ I found out that $$\left|\sum_{n=1}^{\infty}(-1)^{n+1}a_n - \sum_{n=1}^{N}(-1)^{n+1}a_n\right| = \left|\sum_{n=N+1}^{\infty}(-1)^{n+1}a_n\right|,$$ but I don't know how to continue that proof.
I would appreciate any help!

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1 Answer 1

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That is not true in general. Let $a_{2i - 1} = \frac1i$ and $a_{2i} = \frac1{2^i}$. Then $$\left|\sum_{n=N+1}^{+\infty}(-1)^{n+1}a_n\right| = +\infty$$ for all $N$, because $$\sum_{i = k}^{+\infty}a_{2i - 1} = \sum_{i = k}^{+\infty}\frac1i = +\infty$$ and $$\sum_{i = \ell}^{+\infty}a_{2i} = \sum_{i = \ell}^{+\infty}\frac1{2^i} = \frac1{2^{\ell - 1}}.$$

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