Question #295143

Give one example and also prove that for the complex series which is



1) Conditionally Convergent.



2) Absolutely Convergent.



3) Both.



4) Neither

Expert's answer

1. The series ∑n=1+∞(−1)nn\sum\limits_{n=1}^{+\infty}\frac{(-1)^n}{n} is conditionally convergent, because its terms are alternating, and their modulus tends to 0 (Leibniz' test for alternating series).

The series is not absolutely convergent, since the series of modules of terms is a harmonic series that diverges.


2. The series ∑n=1+∞1n(n+1)\sum\limits_{n=1}^{+\infty}\frac{1}{n(n+1)} is absolutely convergent, because its terms are positive and the sequence of the partial sums converge to 1:

SN=∑n=1N1n(n+1)=∑n=1N(1n−1n+1)=S_N=\sum\limits_{n=1}^{N}\frac{1}{n(n+1)}=\sum\limits_{n=1}^{N}(\frac{1}{n}-\frac{1}{n+1})=

=(1−12)+(12−13)+⋯+(1N−1N+1)=1−1N+1→1=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\dots+(\frac{1}{N}-\frac{1}{N+1})=1-\frac{1}{N+1}\to 1


3. The example from part 2 is the series that is both absolutely and conditionally converges, because any absolutely convergent series is also simply (conditionally) convergent.


4. The series ∑n=1+∞(−1)n\sum\limits_{n=1}^{+\infty}(-1)^n is neither absoilutely or conditionally convergent.

Really, since its terms do not tend to 0, the series cannot be divergent.


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