If $a_j\ge 0$ and $\sum_{j=1}^\infty a_j<\infty$, show $\liminf j a_j=0$ 
Suppose that $a_j\ge 0$ and that $\sum_{j=1}^\infty a_j<\infty$.
(1) Show that $\liminf j a_j=0$.
(2) Give an example showing that $\limsup ja_j>0$ is possible.

How to do this question? I don't know how to understand $\sum_{j=1}^\infty a_j  <\infty$. Suppose one sequence is $-1,1,-1,1,...$; can I say $\sum_{j=1}^\infty a_j <\infty$? Since the sum is either $-1$ or $0$.
Is $\sum_{j=1}^\infty a_j  <\infty$ equivalent to saying the sequence $\{a_j\}$ convergent?
The definition of limsup and liminf I learned is, say, $a_j=(-1)^j$, then to find lim inf $a_j$, then $A_1=\{\inf a_1,a_2,a_3,...\}=\inf\{-1,1\}=-1$
$A_n=\inf\{a_n,a_{n+1},...\}$
Then lim inf $a_j$ is the limit of $A_n$.
 A: $a_j \geq 0$ for all $j$ makes things a little simple for us. This is because, if $a_j \geq 0$ , then we know that $\sum_{j=1}^n a_j$ is a monotone increasing sequence. Therefore, unlike the sequence $-1,1,-1,1,-1,1$ and so on, this sequence doesn't face the ambiguity issue that goes like : what if $\sum_{j=1}^n a_j$ is bounded but does not converge? You gave an absolutely perfect example of this, because $-1,0,-1,0,-1,0$ stayed bounded but did not converge. However, this situation doesn't appear here, because an increasing sequence either diverges to positive infinity or converges to a number.
With that in mind, $\sum_{j=1}^\infty a_j < \infty$ means exactly that the sum is a convergent sum, rather than a sum that just stays bounded. Your effort with the counterexample was appreciated, however!

See, here is a great way of thinking about limsup/liminf problems. I will put it in a nutshell without mathematics, and then tell you how it would work when applying it to a question.

The $\limsup$ and $\liminf$ of a sequence can be controlled by changing just one subsequence of that sequence. In other words, the $\limsup$ and $\liminf$ record the extreme subsequences of the given sequence.

What does this mean? Well, the $\limsup$ and $\liminf$ are limits of sequences found by taking the sup/inf of the sequence, at each index, of all terms after that index. Therefore, the idea is that controlling the $\limsup$ and $\liminf$ of a sequence amounts to saying that even along any particular subsequence, the sequence cannot behave too badly (in terms of growing too large,too small etc. as the $\limsup$ dictates).
The difference between the $\limsup$ and the $\lim$ alone, is that the $\lim$ need not exist, but the $\limsup$ can exist, for example for the sequence $-1,1,-1,1,...$. The point is that the subsequence $1,1,1,1,1,...$ decides the $\limsup$ of the sequence, so a particular subsequence affects the $\limsup$.

With this, let's come to our problem. We have to prove that $\liminf ja_j = 0$. What does this mean? This means, that any particular subsequence $n_k a_{n_k}$ cannot run off : we must have, for any subsequence $n_k$ that $n_ka_{n_k}$ converges to $0$ as $k$ converges to infinity.
But let's do it in words first. Let's imagine that the $\liminf$ is bigger than $0$, say it is $C$. This means, that even the slowest growing subsequence of the sequence is growing quite fast. Fast enough for $\sum_{j=1}^\infty a_j < \infty$ to break. How are we going to break it? Well, $ja_j>C$ implies that $a_j > \frac{C}{j}$. So $a_j$ is going to grow faster than the harmonic series, which is a contradiction. But of course we haven't been formal here!
Now, to the formal proof. Suppose that $\liminf ja_j > 0$. Let $\liminf ja_j = C$. Then, we know that $\lim_{n \to \infty} A_n = C$ where $A_n = \inf\{na_n,(n+1)a_{n+1},...\}$.
By the definition of the limit existing ,there is an $N>0$ such that $A_n > \frac C2$ for all $n \geq N$.  Now, what does that mean? Well, in particular, $A_N > \frac C2$. So, $$
\inf\{Na_N,(N+1)a_{N+1},...\} > \frac C2 \implies a_n \geq \frac{C}{2n} \forall n \geq N
$$
That's amazing ; we managed to prove that after some time, every term dominates another term dependent on the harmonic series. Now we return to the assumption $\sum_{j=1}^\infty a_j < \infty$ and see how it gets contradicted. For any $n>N$ we have :
$$
\sum_{j=1}^n a_j = \sum_{j=1}^N a_j + \sum_{j=N+1}^{n} a_j \geq \sum_{j=1}^N a_j  + \frac{C}{2} \sum_{j=N+1}^{n} \frac 1{j}
$$
Now, the left hand side has a limit as $n \to \infty$, therefore the right side should as well. But we know that's not true : $\sum_{j=N+1}^{n} \frac 1{j}$ is just the harmonic series after a certain point, and we know that the harmonic series doesn't converge. Therefore, we get a contradiction, and it follows that $\liminf ja_j = 0$.
For the sake of a good life, let's put this neatly for you. Let's write : $$
ja_j = \frac{a_j}{\frac 1j}
$$
So $ja_j$ is like comparing the ratio of $a_j$ to that of the harmonic series term, which is divergent. Thus, the point of the exercise is to tell you this : the terms of a convergent series must , at least along a subsequence, be well below the harmonic series.

Let's now get to the second one. The idea here is quite simple : changing stuff along a subsequence will ensure that the $\limsup$ increases, but if we choose that subsequence carefully then we can succeed in keeping $\sum_{j=1}^\infty a_j$ finite.
What is the bottom line of what we are going to do, in words? Let's look at $ja_j$. If I wanted the $\limsup$ of this sequence to be greater than $0$, then all I need to do, is ensure that along a particular subsequence, this is the case. Nobody cares about the rest of the sequence, as long as convergence is ensured. But now, we are in a great spot : just fill in zeros!
So the idea is :

*

*Along a particular subsequence , we ensure that the tailoring is done carefully, so that along that subsequence $n_k$ we have $n_ka_{n_k} > 0$.


*The rest of the terms are set to $0$.
Well, let's take a nice convergent sequence for starters. I love geometric series, so let's take the series $\frac 12,\frac 14,\frac 18,\frac 1{16},...,\frac{1}{2^n}$. This sums to $1$, as we all know.
We define the sequence :
$$
a_k = \begin{cases}
\frac{1}{k} & \text{k is a power of } 2 \\
0 & \text{otherwise}
\end{cases}
$$
So $a_k$ looks like :
$$
1,\frac 12, 0,\frac 14,0,0,0,\frac 18,0,0,0,0,0,0,0,\frac 1{16},...
$$
So what have we done? We've basically "delayed" the convergent series we had, just so that the limsup condition is satisfied.
There's no doubt that $\sum_{j=1}^\infty a_j <\infty$, after all it's pretty much the geometric series. I leave you to prove this rigorously if you like. (NOTE : PROOF ATTACHED BELOW)
But now, let's look at the $\limsup ja_j$. For that , we have to look at the sequence $ja_j$ itself. But a beautiful pattern reveals itself when we see this sequence :
$$
1,1,0,1,0,0,0,1,0,0,0,0,0,0,0,1,...
$$
so, the sequence $ja_j$ is $1$ at all powers of $2$, and zero elsewhere!
You can then see, of course, that $A_j = \sup\{ja_j,(j+1)a_{j+1},...\}$ will be $1$ for all $j$, so $\lim A_j = 1$, hence $\limsup ja_j = 1$!

So what do we take away from all this? Basically , $\limsup$ and $\liminf$ behaviour are subsequential behaviour trackers. These two guys are kind of "far more lenient" than a limit, because you can edit just a subsequence of a given sequence , to ensure that the sequence's $\limsup$ or $\liminf$ changes.
This notion allows us to then look at the "worst" and "best" possible growth behaviours of a sequence, and then allows us to come up with useful intuitive statements like the first exercise which can be explained without a word of mathematics very beautifully. That's the take away from this problem that should remain.

Let $a_k = \frac {1}{k}$ if $k$ is a power of $2$, and $0$ otherwise. We want to show that $\sum_{j=1}^{\infty} a_j < \infty$. To see this, we need to study the partial sums $\sum_{j=1}^n a_j$. We will the use the following fact :

A bounded monotonic sequence converges.

Indeed, the proof of the above comes quite simply from proving that the supremum of the elements of the sequence is  bounded, and in fact the limit of the sequence as well.
The given sequence $\sum_{j=1}^n a_j$ is monotonic as the $a_j$ are non-negative. The only thing we need to prove is boundedness, but then observe for any $n$ that  if $m = 2^k$ is any power of $2$ bigger than $n$, we have :
$$
\sum_{j=1}^n a_j \leq \sum_{j=1}^m a_j = 1+\frac{1}{2} + ... + \frac 1{2^k} \leq \sum_{l=1}^\infty \frac{1}{2^l} = 1
$$
so the sequence is bounded, hence it converges.
