This question seems basic but I couldn't find an answer :
Let $a_n$ be a decreasing sequence of positive real numbers such that the sum $$\sum_{i=1}^{\infty} a_i$$ converges to $a$ . Then is it true that there is some $b$ with $0<b<a$ such that for every subsequence $b_n$ of $a_n$ the sum $$\sum_{i=1}^{\infty} b_i$$ doesn't converges to $b$ ?
I thought about it a little and I think the answer is yes but I couldn't find a proof .
My approach
Choose the terms of the sequence $b_n$ in a greedy way :
If you have already chosen $b_1,b_2,\ldots b_{n-1}$ then choose $b_n$ the largest number such that $$\sum_{i=1}^{n} b_i< b$$ Then I think that at some point even if you'll choose all the numbers remaining in $a_n$ you won't make it to the number $b$ .
This is true for a 'fast' decreasing sequence like $\frac{1}{n^2}$ :
$$1+\frac{1}{2^2}+\frac{1}{3^2}+\ldots=\frac{\pi^2}{6}$$
Let's take $b=0.9$ . It's obvious that you can't choose $1$ and all the other terms together have a sum of $\frac{\pi^2}{6}-1<0.9$ so $b=0.9$ won't work .
This can be generalized for all sequences with $a_1 > a-a_1 $ .
This is all I found . Thanks for your help with this question.
