Is there a series $\sum (a_n) $ that converges conditionally but $\sum (a_n -1/n) $ doesn't? I'm studying for a test in calculus and have encountered a question I can't find a proof that contradicts the existence of such series.
Contradict the existence of the series such that:


*

*$\sum(a_n) $ that converges conditionally but $\sum(a_n -1/n) $ doesn't absolutely converge?

*$\sum(a_n) $ that converges conditionally but $\sum(a_n -1/n^2) $ doesn't absolutely converge?


Any help will be greatly appreciated.
Thank you very much
 A: With $a_n=0$, $\sum a_n$  converges in every sense, $\sum(a_n-\frac1n)$ diverges.
With $a_n=(-1)^n\frac1n$, $\sum a_n$ converges conditionally, $\sum (a_n-\frac1{n^2})$ converges conditionally, but not absolutely.
A: I hope the below is what you asked for; it is based on the assumption that you asked to refute existence of conditionally, but not absolutely, convergent series $\sum a_n$ with the stated properties. 

Unfortunately, both of your statements are incorrect.
Ad 1: Let $a_n = \dfrac{(-1)^n}n$. Then $\sum a_n$ converges conditionally; however it is clear that $\sum a_n - \dfrac1n$ does not converge at all.
Ad 2: Suppose $\sum a_n$ converges conditionally. Then since $\sum \dfrac1{n^2}$ converges absolutely, we have by the sum rule for limits (and hence, series) that $\sum a_n - \dfrac1{n^2}$ will converge conditionally as well. 
Moreover, suppose $\sum a_n - \dfrac1{n^2}$ converged absolutely. Then by the triangle inequality:
$$|a_n| = \left|a_n - \frac1{n^2} +\frac1{n^2}\right| \le \left|a_n-\frac1{n^2}\right| + \frac1{n^2}$$
and since by assumption, the right-hand side is summable, so is the left-hand side; but this implies $\sum a_n$ converges absolutely.
