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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.

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marked as duplicate by Crostul, tired, E. Joseph, Martin Sleziak, Stefan Mesken Dec 1 '16 at 12:38

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    $\begingroup$ 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 $\endgroup$ – reuns Dec 1 '16 at 9:24
  • $\begingroup$ @Crostul: this question is about the ordinary vs. the alternating series. $\endgroup$ – Yves Daoust Dec 1 '16 at 9:35
  • $\begingroup$ in the alternating version positive and negative terms compensate...therefor convergence $\endgroup$ – tired Dec 1 '16 at 10:10
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    $\begingroup$ This is not a duplicate, read the question ! $\endgroup$ – Yves Daoust Dec 1 '16 at 13:04
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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

enter image description here

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 $\log2\approx0.6931$).

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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 $\ln(2)$.

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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

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