Proving that $ \sum_{k>1} \frac{n_{k}-n_{k-1}}{n_k} $ is never finite Is it true that for every strictly increasing sequence $ (n_k)_{k\ge 1} $ of natural numbers, the sequence $ \frac{n_k - n_{k−1}}{n_k} $ is never summable? 
A proof is easy in the case when $ \limsup_{k\to\infty} \frac{n_{k-1}}{n_k} < 1 $ simply because then the individual terms don't converge to 0, but when the $ \limsup $ equals 1, how does one proceed? 
 A: You don't need to assume the sequence consists of integers. Monotonicity and divergence to $+\infty$ will suffice.


(Lemma) If $a_n,b_n \geq 0$ such that $\sum a_nb_n$ converges and $a_n$ is decreasing to 0, then $\lim_{n\to\infty} a_nB_n = 0$, where $B_n$ is the partial sum of $b_n$.

Summation by part gives
$$\begin{aligned}
\sum_{n=M+1}^{N} a_n b_n &= a_N B_N - a_M B_M +\sum_{n=M}^{N-1} B_n (a_n - a_{n+1}) \\
&\geq a_N B_N - a_M B_M + B_M \sum_{n=M}^{N-1} (a_n - a_{n+1}) \\ 
&= a_N B_N - a_M B_M + B_M (a_M - a_N) = a_N (B_N-B_M)
\end{aligned} $$
Since $\sum a_nb_n$ converges, for $\varepsilon > 0$, select $N_0$ such that $N,M \geq N_0$ implies $\sum_{n=M+1}^{N} a_n b_n < \varepsilon$. Fix $M = N_0$ and since $a_n \to 0$, select $N_1$ such that $N\geq N_1$ implies $a_N B_M <\varepsilon$. Hence when $N\geq \max\{N_0,N_1\}$, we have $a_N B_N < 2\varepsilon$. The proof is completed.

Assume your series converges, the lemma says $$\lim_{k\to\infty} \frac{n_k - n_1}{n_k} = 0$$ a contradiction, the limit should be $1$.
A: It's well known that if $0<a_k<1,$ then
$$\tag 1 \prod_{k=1}^\infty(1-a_k) > 0 \iff \sum_{k=1}^{\infty} a_k < \infty.$$
In this problem we can take
$$a_k = \frac{n_k-n_{k-1}}{n_k}=1-\frac{n_{k-1}}{n_k}$$
for $k>1.$ We have
$$\prod_{k=1}^\infty(1-a_k)=\lim_{K\to \infty}\frac{n_1}{n_2}\frac{n_2}{n_3}\cdots \frac{n_{K-1}}{n_K} = \lim_{K\to \infty}\frac{n_1}{n_K} = 0.$$
Thus by $(1),$  $\sum_{k=1}^{\infty} a_k = \infty.$
A: $$
\begin{align}
\sum_{k=1}^\infty\frac{n_k-n_{k-1}}{n_k}
&\ge\sum_{k=2}^\infty\left(1-\frac{n_{k-1}}{n_k}\right)\tag1\\
&\ge\min_{k\ge2}\left(\frac{n_{k-1}}{n_k}\right)\sum_{k=2}^\infty\left(\frac{n_k}{n_{k-1}}-1\right)\tag2\\
&\ge\min_{k\ge2}\left(\frac{n_{k-1}}{n_k}\right)\sum_{k=2}^\infty\log\left(\frac{n_k}{n_{k-1}}\right)\tag3\\
&=\min_{k\ge2}\left(\frac{n_{k-1}}{n_k}\right)\lim_{k\to\infty}\log\left(\frac{n_k}{n_1}\right)\tag4
\end{align}
$$
Explanation:
$(1)$: leave out the $k=1$ term$\vphantom{\lim\limits_{k\to\infty}}$
$(2)$: since $\frac{n_{k-1}}{n_k}\le1$ and $\lim\limits_{k\to\infty}\frac{n_{k-1}}{n_k}=1$, $\min\limits_{k\ge2}\left(\frac{n_{k-1}}{n_k}\right)\gt0$ exists
$(3)$: $x\ge\log(1+x)$$\vphantom{\lim\limits_{k\to\infty}}$
$(4)$: telescoping series
A: If $n_k = a$ for any $a \neq 0$ then convergence follows trivially.
Otherwise, notice that the sequence $(|n_k|)_{k\geq 1}$ can only have $|n_{k+1}| > |n_k|$ finitely many times (why?). However, for any initial choice of $n_0$ there are only finitely many values less than $|n_0|$, so the first sentence implies that the sequence in question is never summable.
(Hint: To show this last piece, one must note not only that there are only finitely many numbers less than $|n_0|$, but that there are only finitely many numbers below the largest number reached by the finitely jumps that the sequence is allowed to take.)
