The series \(\sum \frac{-1}{n}\) is divergent, paralleling the divergence of the harmonic series \(\sum \frac{1}{n}\). This conclusion is drawn from the fact that if the partial sums of the harmonic series diverge, then the partial sums of the negative series also diverge. The negative sign does not affect the divergence property of the series, as it merely reflects the values without altering their behavior towards infinity.
PREREQUISITESStudents of mathematics, particularly those studying calculus and series, educators teaching convergence concepts, and anyone interested in the behavior of infinite series.
\displaystyle (-1)\cdot\sum_{n=1}^\infty \left(\frac{1}{u}\right)=\sum_{n=1}^\infty \left(-\frac{1}{u}\right)haha1234 said:Homework Statement
I know that the harmonic series is divergent.But why \frac{-1}{n} is also divergent?
I've search for some test to test that, but I could not find a method for negative series.
So how can I prove the series is divergent?
Homework Equations
The Attempt at a Solution
haha1234 said:Homework Statement
I know that the harmonic series is divergent.But why \frac{-1}{n} is also divergent?
I've search for some test to test that, but I could not find a method for negative series.
So how can I prove the series is divergent?
Homework Equations
The Attempt at a Solution