If $\lim \limits_{n \to \infty} x_n + x_{n+1} =0 $ is $\lim \limits_{n \to \infty} \frac{x_n}{n}=0$? If $\lim \limits_{n \to \infty} x_n + x_{n+1} =0 $ is $\lim \limits_{n \to \infty} \frac{x_n}{n}=0$?
If $\lim \limits_{n \to \infty} x_n - x_{n+1} =0 $ then I would say $x_n$ is bounded therefore true,
but I have no idea where to start. Can anyone help me with this?
 A: The hint by @Mindlack is half the solution. For $y_n=(-1)^n x_n$, we have $$\lim_{n\to\infty}(y_{n+1}-y_n)=0\implies\lim_{n\to\infty}\frac{y_n}{n}=0,$$ the implication is by the Stolz–Cesàro theorem. Thus, $\lim\limits_{n\to\infty}x_n/n=0.$
A: Consider a sequence built up of x(n) and then construct another one with the rule
x(n)=x(n) for n dividable by 2
x(n)=-x(n) for n not dividable by 2

x(0)+x(1)=x(0)-x(0)=0

x(2n)+x(2n+1)=x(2n)-x(2n)=0

x(2n+1)+x(2n+2)=x(2n+1)-x(2n+1)=0

So the limit of the derived third sequence x(n)+x(n+1) is fo sure zero since all individual members in the sequence are zero.
But our introductory sequence has only the attribute to be a sequence. We just used the rule to construct new sequences from one, particularly selected sequence.
We too used the axiom the indices are unlimited and the mightiness of integer used to enumeration is aleph0.
The first part of our rule set can be renumerated. Now extent the attributes of our sequences:
(a) the sequence is unbounded without any limitation to positive infinity, can be negative infinity too.
(b) the sequence is bounded.
(c) the sequence is convergent.
In case (a) we will find an M in the integers that are always bigger than an n we can select and the magnitude of the sequence element will exceed our bounds.
In cases (a) and (b) we select an n from the indices and name that n. For that n we can find from the definition of convergent sequences that there exists an epsilon positive such that our sum is smaller than that epsilon.
Consider the cases:
(i)   x(n)<0, x(n+1)>0 then x(n+1)>x(n)+x(n+1)
(ii)  x(n)>0, x(n+1)>0 then x(n)+x(n+1)>x(n)>0
(iii) x(n)>0, x(n+1)<0 then x(n)>x(n)+x(n+1)
(iv)  x(n)<0, x(n+1)<0 then x(n)+x(n+1)<x(n)<0

But we like a bound for x(n)/n. Because n in positive integer in our enumeration
the product or quotient has the sign of x(n)
x(n)>0 => x(n)/n>0
x(n)<0 => x(n)/n<0

In the cases (ii) and (iv) we are ready, the to be proven lemmata is correct by the majorant or upper bound criteria.
In the two other cases, we have a lower limit.
There is usually a criterion about absolute convergence in such courses. So we just consider the absolute sequence of the given sequence |x(n)|. Then these two cases lead to the already considered and proven cases. We just need to know, that if a sequence is an absolute convergent than it is convergent without the absolut function too.
The question shows that the requirement for a sequence to have lim x(n)+x(n+1) = 0 is just only as strong as lim x(n)/n = 0 and not as erroneously assumed as strong as lim x(n) = 0 and there is range in between depending on the sequence under consideration. We can prove something in the inverse direction, from lim x(n)/n does not follow that lim x(n)+x(n+1) = 0 in general.
Example for the divergent case:
x(n)=+1/(2) for n dividable by 2
x(n)=−+1/(2) for n not dividable by 2

x(2n)+(2n+1)=2n+1/(4n)-(2n+1)+1/(4n)=-1+1/(2n)
x(2n+1)+(2n+2)=-2n-1+1/(2(2n+1))+(2n+2)+1/(2(2n+2)=1+1/(4n+2)+1/(4n+4))
The sequence condition is not fulfilled because the first part sequence converges to -1 and the second one to 1, but both converge. So things can be rather complicated too for this.
