Prove $0$ is a partial limit of $a_n$

Looking for guidance for the following question:
Let $a_n$ a sequence, such that $\left| {{a_{n + 1}} - {a_n}} \right| < \frac{1}{n}$ and $\{5,-5\}$ are partial limits of $a_n$.
Show that $0$ is also a partial limit.

I don't see how both conditions can live together. Obviously, the difference between the terms of $a_n$ are getting closer and closer, so how can this sequence have $\{5,-5\}$ as partial limits?

I'd be glad for clarification and guidance :)

Solution 1:

You need to show that the exists a subsequence of $\{a_n\}$ converging to zero.

Equivalently, you need to show that, for every $N>0$ and $\varepsilon>0$, there exists an $n>N$, such that $$ |a_n|<\varepsilon. $$ Let $N>0$ and $\varepsilon>0$. We shall find an $a_m$ absolutely smaller that $\varepsilon$, for $m>N$.

As there exists a subsequence of $\{a_n\}$ converging to $5$, the exists $$ n_1>\max\{\lfloor \varepsilon^{-1}\rfloor, N\}, $$ such that $a_{n_1}>0$, and as there exists a subsequence of $\{a_n\}$ converging to $-5$, the exists $$ n_2>n_1, $$ such that $a_{n_2}<0$.

So we have that $a_{n_2}<0<a_{n_1}$.

Let $$ m=\min\{ n: n>n_1 \,\,\&\,\, a_n<0\}. $$ This means that $$ a_m<0\le a_{m-1}, $$ and at the same time $$ |a_m-a_{m-1}|<\frac{1}{m-1}<\frac{1}{n_1}<\frac{1}{\lfloor \varepsilon^{-1}\rfloor}\le \varepsilon. $$ Thus $$ |a_m|=-a_m\le a_{m-1}-a_m<\varepsilon. $$

Solution 2:

The basic reason is that the harmonic series diverges. You can go an infinite distance by using only the steps $s_n := a_{n+1} - a_n$ such that $|s_n| = \frac{1}{n}$. (So, by telescoping series, we have then that $a_n = -a_1 + \sum_{k=1}^{n} s_n$.)

By the way, I have never heard of the term "partial limit", but the definition given in the comment by DonAntonio sounds reasonable though: a partial limit of a sequence $\{a_n\}$ is a limit of a subsequence of $\{a_n\}$. (I would have described this as a limit point of the set of points which comprise the sequence.)