Proving that $\sum_{k=1}^{\infty}\frac{1}{k}$ diverges I am trying to practice proofs that involve the convergence of infinite series. Would someone mind helping explain how the value of $\epsilon = \frac{1}{4}$ is obtained in the following proof ? 
 
I would greatly appreciate some help as I am quite unsure of obtaining the value of $\epsilon$ so that it provides a contradiction. I know that it is from working with the inequality $|\frac{1}{k}-L| < \epsilon$
Thank you
 A: Simple way
Let $\sum_{k=1}^{\infty}\frac 1k=L$ where $L\in \mathbb{R}$. We have
$$L=1+\frac 12+\frac 13+\frac 14+\frac 15+\frac 16+\cdots$$
$$L>\frac 12+\frac 12+\frac 14+\frac 14+\frac 16+\frac 16+\cdots$$
$$L>1+\frac 12+\frac 13+\cdots$$
$$L>L$$
It is contradiction.
A: The choice of $\epsilon = \frac{1}{4}$ is simply the largest one that works in for this particular line of reasoning.
Given that $s_{2n} > s_{n} + \frac{1}{2}$, what is an epsilon where $|s_{n} - L| < \epsilon$ for all large $n$ makes the last line work?
$$\epsilon > s_{2n} - L > s_n + \frac{1}{2} -L = (s_n - L) + \frac{1}{2} > -\epsilon + \frac{1}{2}$$
Taking the left and right parts: $\epsilon > -\epsilon + \frac{1}{2}$ implies
$$2 \epsilon > \frac{1}{2}$$
therefore any positive $\epsilon \le \frac{1}{4}$ will make this a contradiction.
A: The trick is to consider doublings of the number of terms.
Assume you computed the sum of the $n$ first inverses, let $$S_n=\sum_{k=1}^n\frac1k.$$
Then the sum of the $2n$ first inverses is such that
$$S_{2n}=S_n+\sum_{k=n+1}^{2n}\frac1k>S_n+\sum_{k=n+1}^{2n}\frac1{2n}=S_n+\frac12.$$
This is the key property: adding $\frac12$ is just enough to shift the interval $(-\frac14,\frac14)$ to the disjoint $(\frac14,\frac34)$ hence the choice of $\epsilon$.

We can repeat the doublings and by induction,
$$S_{2^kn}=S_n+\frac k2$$ can be made as large as you want.
This also shows that with $m=2^kn$,
$$S_m>S_n+\frac12\log_2\left(\frac mn\right).$$
For large $n$, the sum is better approximated by an integral and
$$\frac1n\sum_{k=n+1}^{2n}\frac nk\approx\int_1^2\frac{dt}t=\log2,$$ so that
$$S_m\approx S_n+\log 2\cdot\log_2\left(\frac mn\right)=S_n+\log\left(\frac mn\right).$$
This can be written
$$S_m-\log m\approx S_n-\log n\approx\gamma$$ where $\gamma$ is a constant known as Euler-Mascheroni.
A: It is because of the specific form of 
$$s_{2n}-s_n = \frac{1}{n+1} + \ldots \frac{1}{2n} > n\frac{1}{2n} = \frac{1}{2}\cdot$$
This inequality is independent of the choice of $\varepsilon$ and comes from the specificity of the harmonic series.
Actually, any $0 < \varepsilon < \frac{1}{4}$ would also work. For example if you took $\varepsilon = \frac{1}{8}$, in the last row you would have obtained $\frac{1}{8} > \ldots  -\frac{1}{8} + \frac{1}{2} = \frac{3}{8},$ which is also a contradiction.
In this case $\varepsilon = \frac{1}{4}$ is the biggest number which contradicts $\varepsilon > \frac{1}{2} - \varepsilon$.
A: We can use the Cauchy condensation test: If $\sum\limits^{\infty}_{n = 1}a_{n}$ is a series in which the terms of the series are positive 
and form a non-increasing sequence, then it converges or diverges with the series $\sum\limits^{\infty}_{k = 1}2^{k}a_{2^{k}}$.
For the harmonic series $\sum\limits^{\infty}_{n = 1} \frac{1}{n}$, using the Cauchy condensation test, $n$ becomes $2^{k}$, so we have 
$$\sum\limits^{\infty}_{k = 1} \frac{2^{k}}{2^{k}} = \sum\limits^{\infty}_{k = 1} 1$$ which clearly diverges. 
Thus, by the Cauchy condensation test, the harmonic series diverges.
A: The key observation is that the sum of the $a_k$ from $k=n+1$ to $k= 2n$ is bounded from below by $1/2$. This value has to exceed  the distance from the lower bound for $s_n -L$ to the upper bound and it will obviously do if this distance is $1/2 = 1/4 - (-1/4)$. Therefore this is a convenient choice. 
Edit actually this observation can easily be used to give a proof not relying on contradiction. You can obviously use this to show that the partial sums will exceed any positive multiple of $1/2$.
A: With $\zeta(x):=\sum\limits_{k=1}^\infty \frac{1}{k^x}$ is $(1-2^{1-x})\zeta(x)=\sum\limits_{k=1}^\infty \frac{(-1)^{k-1}}{k^x}$ convergent for $x>0$ because of the Leibniz criteria. But $\frac{1}{1-2^{1-x}}$ is divergent for $x\to 1$ and therefore $\zeta(1)$.
