# Show that the sequence converges

1. Dec 24, 2013

### ianwood

Using the steps below, show that the following sequence converges:
$1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\frac{1}{10}+\frac{1}{11}-\frac{2}{12}++-++-...$

i. Consider the subsequence (s2,s3,s5,s6,s8,s9,...) of the sequence of partial
sums. Show that this is the sequence of partial sums of a related convergent
series.

ii. Show that the original series is also convergent.

I have tried and showed part ii successfully, by considering
$\frac{1}{2}-\frac{2}{3}+\frac{1}{5}-\frac{2}{6}+\frac{1}{8}-\frac{2}{9}+\frac{1}{11}-\frac{2}{12}+...$
$=(\frac{1}{2}+\frac{1}{5}+\frac{1}{8}+\frac{1}{11}+...)-(\frac{2}{3}+\frac{2}{6}+\frac{2}{9}+\frac{2}{12}+...)$
$=\sum\limits_{k=1}^\infty \frac{1}{3k-1}-\sum\limits_{k=1}^\infty \frac{2}{3k}$

and considering
$1+\frac{1}{4}+\frac{1}{7}+\frac{1}{10}+...$
$=\sum\limits_{k=1}^\infty \frac{1}{3k-2}$

So the sequence =$\sum\limits_{k=1}^\infty \frac{9k-4}{3k(3k-2)(3k-1))}$
is convergent by comparison test
However I am wondering what part (i) is asking. I think (s2,s3,s5,s6,s8,s9,...) is divergent. How can I relate to a convergent series?

Last edited: Dec 25, 2013
2. Dec 24, 2013

### Dick

You can't regroup into three divergent series and then resum them to get a convergent series. The series isn't absolutely convergent and that doesn't have anything to do with the partial sums. Group the series into sums of three consecutive terms, then you can make some argument about the limit of the partial sums.

Last edited: Dec 24, 2013
3. Dec 25, 2013

### ianwood

Is this something I am looking for?

(i) Considering $s3,s6,s9,...$

$$s_{3k} = 1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\frac{1}{10}+\frac{1}{11}-\frac{2}{12}+...+\frac{1}{3k-2}+\frac{1}{3k-1}-\frac{2}{3k}$$
$$=\sum\limits_{k=1}^\infty (\frac{1}{3k-2} + \frac{1}{3k-1} - \frac{2}{3k})$$
$$= \sum\limits_{k=1}^\infty \frac{9k-4}{3k(3k-2)(3k-1))}$$

and I can prove convergence by comparison test.

Similarly, considering $s2,s5,s8,...$
$$s_{3k+2} = 1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\frac{1}{10}+\frac{1}{11}-\frac{2}{12}+...-\frac{2}{3k}+\frac{1}{3k+1}+\frac{1}{3k+2}$$

$$=1+\frac{1}{2}-\sum\limits_{k=1}^\infty (\frac{2}{3k} + \frac{1}{3k+1} +\frac{1}{3k+2})$$
$$= \frac{3}{2}-\sum\limits_{k=1}^\infty \frac{9k+4}{3k(3k+1)(3k+2))}$$

and I can prove convergence by comparison test.

Last edited: Dec 25, 2013
4. Dec 25, 2013

### haruspex

You may be misinterpreting that notation. sn here is the sum of the first n terms of the original sequence. That does produce a convergent sequence.

5. Dec 25, 2013

### Dick

Yes, the sequence {s3,s6,s9,...} must converge to some limit L by your comparison test. Now all you need to say is that the limit of {s1,s4,s7,...} and {s2,s5,s8,...} must converge to the same limit. The difference between $s_{3n}$ and $s_{3n-1}$ and $s_{3n-2}$ becomes vanishing small as n->infinity, doesn't it?