What does the sum notation $\lim_{n\to\infty} \sum_{j=n}^\infty a_j$ mean?

Question

What does the sum notation$$\lim_{n\to\infty} \sum_{j=n}^\infty a_j$$ mean?

Is the $n$ fixed or not?

Answer 1

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 $\lim_{n\to\infty}\sum_{j=n}^\infty a_j = 0$.

Answer 2

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 $-1/2$)