Show that the sequence converges

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?

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Homework Helper
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?

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.

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ianwood
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.

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.

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Homework Helper
Gold Member
I think (s2,s3,s5,s6,s8,s9,...) is divergent.
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.

(i) Considering $s3,s6,s9,...$
$$s_{3k} = 1+\frac{1}{2}-\frac{2}{3}+\frac{1}{4}+\frac{1}{5}-\frac{2}{6}+\frac{1}{7}+\frac{1}{8}-\frac{2}{9}+\frac{1}{10}+\frac{1}{11}-\frac{2}{12}+...+\frac{1}{3k-2}+\frac{1}{3k-1}-\frac{2}{3k}$$
$$=\sum\limits_{k=1}^\infty (\frac{1}{3k-2} + \frac{1}{3k-1} - \frac{2}{3k})$$
$$= \sum\limits_{k=1}^\infty \frac{9k-4}{3k(3k-2)(3k-1))}$$