Cfrgtkky

In my real analysis exam I had a problem in which I proved that $|x_{n+1} - x_n|lt {a^n}$ for all natural numbers $n$ and for all positive number $alt 1$ 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?

Consider the harmonic series: $sum_{n=1}^{infty}frac 1n$.

$mid a_{n+1}-a_nmid=frac1{n+1}ltfrac1n$.

But it diverges.

No, $lvert x_{n+1}-x_nrvert < 1/n$ does not imply that $(x_n)_{nin mathbb{N}}$ is Cauchy. Consider $x_n = sum_{k=1}^n 1/2k$, which does not converge.

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.

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.

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.

This would imply that any series with a general term which tends to $0$ is convergent. This is false (except for $p$-adic numbers…).