Why does $\sum_{k=1}^{\infty} \frac {(-1)^{k-1}}{k}$ converge conditionally?

Question

Why does it converge conditionally? $$\sum_{k=1}^{\infty} \frac {(-1)^{k-1}}{k}$$

Answer 1

The given series is convergent by the Leibniz's theorem but it isn't absolutely convergent since the harmonic series is divergent hence this series is conditionally convergent.

Answer 2

The series $\sum_{k=1}^{\infty} \frac {(-1)^{k-1}}{k}$ converges conditionally because it fulfills the Leibnitz Theorem ($\frac{1}{k}$ is a decreasing zero-sequence) but does not converge absolutely because $\sum_{k=1}^{\infty} \frac {1}{k}$ diverges. To understand why your series converges, you can maybe write the first terms out: $$\sum_{k=1}^{\infty} \frac {(-1)^{k-1}}{k} = 1-\frac {1}{2} +\frac {1}{3}-\frac {1}{4}...$$ With this you can see that the $(-1)^{k-1}$ is necessary for convergence, because it "makes the sum smaller".

Answer 3

This is a direct argument that this series sometimes converges and sometimes does not, depending on how you order the sum. The more standard thing to do is to define "conditionally convergent" as a series that converges, but not absolutely. In that case it is clear that this series converges by the alternating series test, and it is clear that it does not converge absolutely because its absolute value is the harmonic series. The following argument does not look at the absolute value of the series at all.

The series $1+\frac13+\frac15+\frac17+\cdots$ diverges, since term by term it is larger than $\frac12+\frac14+\frac16+\frac18+\cdots$, which is one half the harmonic series.

So it's always possible to add together enough consecutive terms from this series to obtain a result larger than any positive number. Start with $1+\frac13+\frac15$, which is larger than $1+\frac12$. Then add only one negative term from the alternating harmonic series:

$$\color{blue}{1+\frac13+\frac15}-\frac12>1$$

At this point we have a sum larger than $1$. Now continue adding positive terms until we reach a sum larger than $2+\frac14$. It is guaranteed we will eventually get there, because of the sum of the odd terms diverging.

$$\color{blue}{1+\frac13+\frac15}-\frac12\color{blue}{+\frac17+\frac19+\cdots+\frac{1}{69}}=2.259\ldots$$

Now we can throw in $-\frac14$, and we are at a grand total $$\color{blue}{1+\frac13+\frac15}-\frac12\color{blue}{+\frac17+\frac19+\cdots+\frac{1}{69}}-\frac14>2$$

Continue in this way, and you can get an ordering with partial sums that exceed any whole integer $n$. (We have exceeded $1$ and $2$ so far.) It will take more and more positive terms, each go around, but will always be doable. Here is the next iteration:

$$\color{blue}{1+\frac13+\frac15}-\frac12\color{blue}{+\frac17+\frac19+\cdots+\frac{1}{69}}-\frac14\color{blue}{+\frac1{71}+\frac1{73}+\cdots+\frac1{709}}-\frac16>3$$

This ordering will include all of the alternating harmonic terms, since it is adding in the odd ones and even ones in sequence (but at separate rates). So we will have built an ordering where the sum diverges.

Note that $1-\frac12+\frac13-\frac14+\cdots$ converges to $\ln(2)$ by several well-known arguments. So in the most basic sense, the alternating harmonic terms sum to a finite total or not depending on what order you sum them in. Hence we have a conditionally convergent series by its more direct meaning. (The indirect meaning is to define a conditionally convergent series as a series that is convergent in one ordering but is not absolutely convergent. It's really more of a theorem though that this definition gives you a "conditionally" convergent series according to the everyday English meaning of "conditionally".)

Answer 4

MrWho, if you're looking for a proof,

If order for the series to converge absolutely, we must have

$$\sum_{k=1}^{\infty} \frac {(-1)^{k-1}}{k}$$ converge and

$$\sum_{k=1}^{\infty} \frac {|(-1)^{k-1}|}{k} = \frac{1}{k}$$

converge as well.

We can use the alternating series test to prove that the first series is convergent.

Let $b_n$ = \frac{1}{k}

Then we must conclude that:

$$\lim_{x -> \infty} b_n = 0$$

Which it does

And also,

$$b_n > b_{n+1}$$

which is true for all values of $b_n$

Thus this series converges by the Alternating Series test.

Now for our second series, use the p-test.

$p =1$ in this case, and whenever $p \le 1$ it diverges.

Because this series diverges, the series is conditionally convergent.