Making sure if it is Cauchy In my real analysis exam I had a problem in which I proved that given any positive number $a\lt 1$ if $|x_{n+1} - x_n|\lt {a^n}$ for all natural numbers $n$ then $(x_n)$ is a Cauchy sequence.
This was solved successfully but the question is if $|x_{n+1} - x_n|\lt \frac 1n$ does that mean $(x_n)$ is Cauchy? Well my answer was yes because I could write this in the form of the first one, but now I am somehow confused with what I have answered since $1/n$ is a sequence of $n$ so maybe the answer is not necessarily true... Can you please provide me with the correct answer for this question?
 A: No, $\lvert x_{n+1}-x_n\rvert < 1/n$ does not imply that $(x_n)_{n\in \mathbb{N}}$ is Cauchy. Consider $x_n = \sum_{k=1}^n 1/2k$, which does not converge.
A: Take $x_n=1+\frac 1 2+\cdots+\frac 1 n$. This is not Cauchy because the harmonic series $1+\frac 1 2+\cdots$ is divergent. 
A: This kind of thing works only if you can show $|x_{n + 1} - x_{n}| < a_n$ where $\sum_{k = 0}^\infty a_k < \infty$ because, if this condition holds,
\begin{align}
|x_{n} - x_{n + m}| &= |x_{n} - x_{n + 1} + x_{n + 1} - x_{n + 2} + x_{n + 2} - \cdots + x_{n + m - 1} - x_{n + m}| \\
&\le |x_{n} - x_{n + 1}| + \cdots + |x_{n + m - 1} - x_{n + m}| \\
&\le a_n + a_{n + 1} + \dots + a_{n + m - 1} \\
&\le \sum_{k = n}^\infty a_k
\end{align}
Now convergence of $\sum a_k$ to $A$ means that for any $\varepsilon > 0$ there exists $N$ such that for all $n \ge N$,
$$ \left| A - \sum_{k = 0}^{n - 1} a_k \right| = \sum_{k = n}^\infty a_k < \varepsilon $$
Comparing this with the above, we have for every $n \ge N$ and every $m \ge 0$,
$$ |x_n - x_{n + m}| < \varepsilon $$
Which means the sequence $(x_n)$ is Cauchy.
If the bound on $|x_{n + 1} - x_n|$ does not converge as a series, you need to use a different trick.
A: For each $n \in \mathbb{N}$, define $x_n:=1^{-1}+2^{-1}+\cdots+n^{-1}$. Notice $|x_{n+1}-x_n|=\frac{1}{n+1}$ but the sequence $\{x_n\}_{n=1}^\infty$ is not Cauchy as its terms are just the partial sums of the harmonic series which is known to not converge and $\mathbb{R}$ is complete.
A: Consider the harmonic series:   $\sum_{n=1}^{\infty}\frac 1n$.
$\mid a_{n+1}-a_n\mid=\frac1{n+1}\lt\frac1n$.
But it diverges. 
A: This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).
