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

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.

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

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.

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.

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