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.

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.

share|improve this question
    
Any sequence $a_n$ converging to zero such that there exists parenthesis with every term inside the parenthesis will work. You could replace your $2^n$ in the denominator by a $\log(n)$ or a $n^n$, it doesn't matter. As long as you use your trick and put enough brackets. =) –  Patrick Da Silva Jun 2 '12 at 16:41
    
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. –  David Mitra Jun 2 '12 at 16:56

2 Answers 2

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.

share|improve this answer

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

share|improve this answer

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.