1
$\begingroup$

Give an example of a positive valued sequence $(a_n)$ such that $\sum a_n$ converges, but $a_n$ may not be splitted as $$a_n=b_n-c_n,$$ where sequences $(b_n),~(c_n)$ are positive valued, decreasing and the series $\sum b_n,~\sum c_n$ converge.

I have not managed to find such a sequence so far: for example, $\dfrac{1}{n^p},~p>1$ does not work, since $\dfrac{1}{n^p}=\dfrac{2}{n^p}-\dfrac{1}{n^p}$.

Thanks for the help.

$\endgroup$
  • 1
    $\begingroup$ You shouldn't take a sequence $(a_n)$ that is decreasing, otherwise $b_n=2a_n$ and $c_n=a_n$ will always work. $\endgroup$ – AlexL Oct 6 '18 at 19:24
3
$\begingroup$

Let $(x_n)$ be any sequence of positive terms with $\sum x_n$ convergent. Define sequence of positive terms $(a_n)$ as follows: $$ a_n=\begin{cases} \frac1{k^2}&\text{if $n=k(k+1)/2$ for some integer $k$} \\ x_n&\text{otherwise} \end{cases} $$ So the first few terms of $(a_n)$ are: $$ \textstyle \frac11,\ x_2,\ \frac14,\ x_4,\ x_5,\ \frac19,\ x_7,\ x_8,\ x_9, \ \frac1{16},\ x_{11},\ x_{12},\ \ldots\tag1 $$ It is clear that $\sum a_n$ converges, since $\sum a_n$ is at most $\sum x_n + \sum_k\frac1{k^2}$.

Now consider any decomposition of the form $a_n=b_n-c_n$ with $(b_n)$ positive and decreasing, and $(c_n)$ positive. We show that $\sum b_n$ cannot converge.

By construction, $b_n> a_n$ for every $n$. In particular, checking the triangular subscripts $n:=k(k+1)/2$, we see $b_1>\frac11$, $b_3>\frac14$, $b_6>\frac19$, $b_{10}>\frac1{16}$. By assumption, $(b_n)$ is decreasing, so the sequence $(b_n)$ must be at least as big (pointwise) as the sequence $$ \textstyle \frac11,\ \frac14,\ \frac14,\ \frac19,\ \frac19,\ \frac19,\ \frac1{16},\ \frac1{16},\frac1{16},\ \frac1{16},\ \frac1{25},\ \ldots\tag2 $$ But the sum of the terms (2) is the harmonic series. Therefore $\sum b_n$ diverges.

$\endgroup$
1
$\begingroup$

If $(a_n)$ is decreasing then, as you pointed it, we can take

$$b_n=2a_n$$ and $$c_n=a_n$$

If $(a_n)$ is increasing, as a positive sequence which converges to zero $(\sum a_n \text{ is convergent})$, it will satisfy $a_n=0$.

$( a_n)$ must alternate.

$\endgroup$
  • $\begingroup$ Thank you. Non-decreasing does not imply increasing though and also we are looking for a positive ($>0$) valued sequence $(a_n)$. $\endgroup$ – Nikolaos Skout Oct 6 '18 at 19:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.