What does the sum notation $\lim_{n\to\infty} \sum_{j=n}^\infty a_j$ mean? What does the sum notation$$\lim_{n\to\infty}  \sum_{j=n}^\infty a_j$$
mean?
Is the $n$ fixed or not?
 A: It is the limit of the sequence
$$\sum_{j=1}^\infty a_j,\;
  \sum_{j=2}^\infty a_j,\;
  \sum_{j=3}^\infty a_j,\;
  \sum_{j=4}^\infty a_j,\;
  \sum_{j=5}^\infty a_j,\;
   \cdots
$$
My first guess would be, if   $\sum_{j=1}^\infty a_j$ exists, then your answer has to be $0$.
For example
$$\lim_{n\to\infty} \sum_{j=n}^\infty \frac{1}{2^j} =
  \lim_{n\to\infty} \frac{2}{2^n} =
  0
$$
THEOREM. If
$\sum_{j=1}^\infty a_j$ exists, then
$\lim_{n\to\infty} \sum_{j=n}^\infty a_j = 0$
PROOF. For convenience, let $\sum_{j=1}^\infty a_j = L$. Then, since the limit exists,
$L = \sum_{j=1}^{n-1} a_j + \sum_{j=n}^\infty a_j$
Now let $n \to \infty$. Then $L = L + \lim_{n\to\infty}\sum_{j=n}^\infty a_j$
So $\lim_{n\to\infty}\sum_{j=n}^\infty a_j = 0$.
A: I would interpret it as $$\lim_{n \to \infty} \lim_{m \to \infty} \sum_{j = n}^m a_j$$
Inside the two limits, both $n$ and $m$ are fixed.You then take the two limits on the expression found.
For example, you will arrive at an expression like $\sum_{j = n}^m a_j = \frac 1m + \frac {2n^2 + 1}{4n^2} - 1$ (notice how it depends only on $n$ and $m$).
Then you take both limits to find the result (in this case $-1/2$)
