The series $\sum_{n=1}^\infty\frac1n$ diverges!

We all know that the following harmonic series

$$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$

diverges and grows very slowly!! I have seen many proofs of the result but recently found the following: $$S =\frac 1 1 + \frac 12 + \frac 13 +\frac 14+ \frac 15+ \frac 16+ \cdots$$ $$> \frac 12+\frac 12+ \frac 14+ \frac 14+ \frac 16+ \frac 16+ \cdots =\frac 1 1 + \frac 12 + \frac 13 +\cdots = S.$$ In this way we see that $S > S$.

Can we conclude from this that $S$ is divergent??

Solution 1:

The proof can be made a bit more rigorous by setting $$ \begin{align} a_n=\frac1n:&\,\quad1,\,\frac12,\frac13,\frac14,\frac15,\frac16,\dots\\b_n=\frac1{2\lfloor(n+1)/2\rfloor}:&\quad\frac12,\frac12,\frac14,\frac14,\frac16,\frac16,\dots \end{align}\tag{1} $$ Note that $a_n\ge b_n$, $a_n\gt b_n$ when $n$ is odd, and $a_n=b_{2n-1}+b_{2n}$.

Assuming that $$ \sum_{n=1}^\infty a_n\tag{2} $$ converges, then $$ \sum_{n=1}^\infty b_n=\sum_{n=1}^\infty(b_{2n-1}+b_{2n})=\sum_{n=1}^\infty a_n\tag{3} $$ also converges. However, $$ \sum_{n=1}^\infty(a_n-b_n)\gt0\tag{4} $$ Since $a_n\ge b_n$ and $a_n\gt b_n$ when $n$ is odd.

Now, $(3)$ says that $$ \sum_{n=1}^\infty b_n=\sum_{n=1}^\infty a_n\tag{5} $$ and $(4)$ says that $$ \sum_{n=1}^\infty b_n\lt\sum_{n=1}^\infty a_n\tag{6} $$ These last two statements are contradictory, so the assumption that $(2)$ converges must be false.

Solution 2:

If $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ both exist, $a_n\ge b_n$ for all $n$, and $a_i>b_i$ for at least one $i$, then the first sum must be strictly greater than the second. This is because the first's partial sum is eventually always at least $a_i-b_i$ more than the second's partial sums. In this case, one can subsequently reason that if the first exists, so does the second. If this is your reasoning, it is valid.

Solution 3:

There's another way to approach this, via integration: compare - on the domain $(1, \infty)$ the curve $y_1={1\over x}$ with the step function $y_2={1\over Floor(x)}$ (where $Floor(x)$ is the greatest integer $<x$).

Clearly, for every $x\in (1,\infty)$, we have $0<y_1(x)\le y_2(x)$, so $\int_1^\infty y_1dx\le\int_1^\infty y_2dx$; moreover, $\int_1^\infty y_2dx$ is just the sum of the harmonic series.

But integrating, we get $\int_1^\infty y_1dx=\ln(x)\vert^\infty_1=\infty$, so the harmonic series must diverge.

Of course, this is non-rigorous, but it's good motivation, and it can be made rigorous without much work.

Solution 4:

Here is another approach to making the answer more rigorous. Assume the series converges, then $$ \begin{align} \sum_{k=1}^\infty\frac1k &=\sum_{k=1}^\infty\left(\frac1{2k-1}+\frac1{2k}\right)\\ &=\sum_{k=1}^\infty\left[\left(\frac1{2k}+\frac1{2k}\right)+\left(\frac1{2k-1}-\frac1{2k}\right)\right]\\ &=\sum_{k=1}^\infty\frac1k+\sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right)\\ &=\sum_{k=1}^\infty\frac1k+\log(2) \end{align} $$ The rearrangements are justified because the series are all of positive terms.

The sum $$ \begin{align} \sum_{k=1}^\infty\left(\frac1{2k-1}-\frac1{2k}\right) &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\\[6pt] &=\log(2) \end{align} $$ is a well-known series for $\log(2)$.