Sum of infinite positive numbers

Question

I have the following sum of infinite positive numbers $$\sum_{k=1}^\infty a_{k}^2$$ and I know that this sum converges, meaning that $$\sum_{k=1}^\infty a_{k}^2 < \infty$$ The book I am studying says that since the above sum converges, the following must be true $$\lim_{k\rightarrow\infty}a_{k} = 0$$

I think that this is true because if I want the sum of infinite positive numbers to be finite, these positive terms have to be close or equal to $0$. However, I'm not sure why it takes the limit, because that implies that as $k$ goes to $\infty$ then the terms start to go close to $0$, correct? I know that the sum of positive numbers is finite then $a_{k}=0$ for all but at most countably many $k$, for which $a_{k}\neq0$.

So, first of all, are my above thoughts correct, and if yes, why does it take the limit and sets it equal to $0$ and doesn't just say that $a_{k}=0$ for all but at most countably many $k$?

Thanks in advance, sorry if my thoughts are a bit jumpled!

Answer 1

Let consider the partial sum

$$S_n=\sum_{k=1}^n a_{k}^2 \to L$$

then

$$a_{n+1}^2=S_{n+1}-S_n \to L-L= 0 \implies a_n \to 0$$

Therefore the necessary condition for convergence is that $a_n$ tends to zero which doesn't imply that eventually $a_n=0$.

Answer 2

You're mixing things up with sequences indexed by uncountable sets.

Only countably many terms may be non-zero for the series to converge (you have a typo here btw, it should read $\neq 0$), but this is only a necessary condition, not sufficient (think of the sequence of countably many $1$'s).

To see that $a_k$ must tend to $0$ as $k$ goes to infinity, consider the difference of two consecutive partial sums (both of which converge to the sum of the series).

Answer 3

I know that the sum of positive numbers is finite then ak=0 for all but at most countably many k, for which ak≠0.

You only have countably many $a_k$. It makes no sense to talk of for all but at most countable because any amount, even absolutely none or never, is "for all but at most countably many"... because the time it does happen (which is always) is only countably many.

Now surely you have come across $1 + \frac 12 + \frac 14 + \frac 18 + .... = 2$.

In this case $a_k = \frac 1{2^{k-1}}$ and $\sum a_k = 2$ and $a_k \ne 0$.

And it is true that $a_k =0$ for all but at most countably many $a_k$ because the $a_k$ where $a_k \ne 0$ are .... all of them... and all of them is at most countably many.