Harmonic Series Convergence


Solution 1

This is a graphical explanation. The partial sums of the harmonic series is given by $$ S_n=\sum_{k=1}^n\frac1k $$ and they look like this

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The partial sums of the alternating harmonic series is given by $$ S_n=\sum_{k=1}^n\frac{(-1)^{k+1}}k $$ and they look like this

enter image description here

These graphs might give us some idea that the first one probably diverges and the second probably converges (actually the limit of the alternating harmonic series is $\log2\approx0.6931$).

Solution 2

Alternating series converge "easier". If you consider the sums of the even and the odd terms separately, they will tend to the same limit (with an added constant). Then if you subtract them, the residue has more chance to converge than the sum.

Graphically, the even and odd sums will look "parallel" to each other as they tend to a constant difference, while both can diverge.

enter image description here

Solution 3

The divergence of the harmonic series is a well known result in basic Mathematical Analysis. Several proofs of this result exists. The theorem can be proved by using the integral test, which may give you a graphical feel. One elementary proof relies on the principle of contradiction: Assume on the contrary, that $\sum_1^\infty \frac{1}{n}$ is convergent and its sum is $L$. Now, $L=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdot\cdot\cdot\implies L=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)+(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...)=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)+\frac{1}{2}(1+\frac{1}{2}+\frac{1}{3}+...)=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)+\frac{L}{2}>L$

because$$(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...)$$ is termwise greater than $$(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...)$$ Thus, no such sum $L$ exists.

As for alternating series, there is a theorem called Riemann Rearrangement Theorem which states that any rearrangement of a conditionally convergent series can be used to produce any sum in $(-\infty, \infty)$. Thus, though the alternating series is convergent, it is conditional and a clever rearrangement of the order of sums can make the series diverge. As for your last statement that the sequence tending to $0$, there are several series having terms tending to zero which diverge.

The alternating series has a sum equal to $\ln(2)$.

A_for_ Abacus
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A_for_ Abacus

Updated on December 03, 2022


  • A_for_ Abacus
    A_for_ Abacus 11 months

    Could someone explain to me graphically how the harmonic series can be divergent while the alternating harmonic series can be convergent since they are both using 1/n in their series and going towards 0.

    • reuns
      reuns almost 7 years
      because $\frac{1}{2n-1}-\frac{1}{2n} = \frac{1}{2n(2n-1)} < \frac{1}{n^2}$ and $\sum_n \frac{1}{n^2}$ converges
    • Admin
      Admin almost 7 years
      @Crostul: this question is about the ordinary vs. the alternating series.
    • tired
      tired almost 7 years
      in the alternating version positive and negative terms compensate...therefor convergence
    • Admin
      Admin almost 7 years
      This is not a duplicate, read the question !