Is there a series $\sum (a_n) $ that converges conditionally but $\sum (a_n -1/n) $ doesn't?

Question

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

Answer 1

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.

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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.

Answer 2

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.