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find an example for a series $a_{n}$ that satisfies the following:

  1. $a_{n}\xrightarrow[n\to\infty]{}0$

  2. ${\displaystyle \sum_{n=1}^{\infty}a_{n}}$ does not converges

  3. There is a way to insert parentheses so ${\displaystyle \sum_{n=1}^{\infty}a_{n}}$ will converges.

I was thinking about the series:$ 1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$

But I don't know how to prove 2.

Also will be nice to hear another examples, if any.

  • 0
    Look at the sequence of partial sums, $(S_n)$, defined by $S_n=\sum_{k=1}^n a_k$. It should be clear how to show that this sequence does not converge (find a subsequence that alternates between $0$ and $1$, e.g.). Recall that an infinite sum converges iff its sequence of partial sums converges.2012-06-02

2 Answers 2

1

The series $1-1+\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}+...$ is indeed divergent, because it has infinitely many partial sums equal to $1$, and infinitely many partial sums equal to $0$. Inserting parentheses as $\left(1-1\right)+\left(\frac{1}{2}+\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}-\frac{1}{4}\right)+...$ yields a convergent series.

More generally: take any divergent series in which $\limsup S_N\ge 0$ and $\liminf S_N\le 0$ ($S_N$ are partial sums). Split every term into several, to ensure $a_n\to 0$. Insert $k$th closing parenthesis when the partial sum drops below $1/k$ in absolute value.

0

D'Alembert's ratio test will help you with nr. 2.