1

Leibniz's convergence test (Theorem $10.14$ in Tom Apostol's Calculus, vol 1) is that if ${a_n}$ is a monotonically decreasing sequence with limit $0$, then the alternating series $\sum_{n=1}^{\infty} (-1)^{n-1}a_n$ converges.

Exercise $10.20.37$ in Tom Aposto's Calculus vol. $1$ is to find all constant complex numbers $z$, such that the series $\sum_{n=1}^{\infty} \frac{(-1)^n}{z+n}$ converges.

When solving that, it would be handy to use a complex analog of Leibniz test (where ${a_n}$ are complex numbers, with limit of 0), but I don't know how to do that, because there is no ordering between complex numbers. Furthermore, it seems like (link) just convergence of $a_n$ is not enough.

What would be a good way forward to solve the exercise?

Thanks!

S11n
  • 978

1 Answers1

1

Hints: If $x=-n$ for some $n \in \mathbb N$ then the series is not defined. Otherwise, there is a positive distance form $-x$ to $n \in \mathbb N$. The given series is $\sum \frac {(x+n)-iy} {(x+n)^{2}+y^{2}}(-1)^{n}$. Check that $\sum \frac {-iy} {(x+n)^{2}+y^{2}}(-1)^{n}$ is absolutely convergent. No apply Liebniz test to $\sum \frac {(x+n)} {(x+n)^{2}+y^{2}}(-1)^{n}$. The terms of $\sum \frac {x+n} {(x+n)^{2}+y^{2}}(-1)^{n} $decrease to $0$ after a certain number of terms and this is enough for the application of Liebniz test. Conclusion: The series converges whenever $-x$ is not in $\mathbb N$,

Kavi Rama Murthy
  • 359,332