Does this condition imply that the sequence is cauchy?

Question

We are given : $|a_n-a_{n+1}|\le \frac{n\text{log}n}{e^n}$, we have to determine whether the sequence $\{a_n\}_{n=1}^{\infty}$ is cauchy or not.

Since,$|a_n-a_m|\leq |a_n-a_{n+1}|+\dots+|a_{m-1}-a_m|$. I believe it is enough to prove that the sequence $\sum_{n=1}^{\infty}\frac{n \text {log}n}{e^n}$ converges.

Can someone please help if the sequence $\sum_{n=1}^{\infty}\frac{n \text {log}n}{e^n}$ converges or not. and is my approach to the above question correct?

Answer 1

Hint: Apply Ratio test to show that $\sum \frac {n \ln n} {e^{n}}$ is convergent. [$lim \frac {a_{n+1}} {a_n}=\frac 1 e <1$ in this case].

Answer 2

The root test also works since $(n\ln(n))^{1/n} \to 1$.

As a matter of fact, every test will work on this since it is quite strongly convergent.

Answer 3

Your approach is correct because you can write $$a_n = a_1 + \sum_{k=1}^{n-1}(a_{k+1}-a_k)$$

So, if you show that $\sum_{k=1}^{\infty}(a_{k+1}-a_k)$ is convergent, then the sequence $a_n$ must be convergent as well and, hence, it is Cauchy.

You can show the (absolute) convergence of the series, for example, by direct comparision:

$$\frac{n\log n}{e^n} \stackrel{\log n < n, e^n =\sum_{k=0}^{\infty} \frac{n^k}{k!}}{\leq} \frac{n^2}{\frac{n^4}{4!}}=\frac{4!}{n^2}$$

Answer 4

$\log n <n$ for $n \ge 1$.

$0< \frac{n\log n}{e^n} <\frac{n^2}{e^n}<\frac{4!}{n^2}$(comparison test).

Recall:

$e^n >$

$1+n+n^2/2!+n^3/3!+n^4/4!+...>n^4/4!.$