$\sum a_n$ converges but $a_n=b_n-c_n$ for suitable $(b_n),(c_n)$ is impossible

Question

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.

Answer 1

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.

Answer 2

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.