# Can a subsequence of a convergence sum converge anywhere?

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.

• @WojciechKarwacki So you mean to take $d$ and $e$ the two biggest elements and then $b=\frac{d+e}{2}$ will satisfy the property ? Why ? – user252450 Jan 3 '16 at 20:25

Let $a_n=\frac{1}{2^n}$. Then for every $b$ with $0\lt b\lt 1$ there is a subsequence of $(a_n)$ whose sum converges to $b$.
• It is worth noting, perhaps, that the property does hold for $a_n = r^{n}$ when $r < 1/2$ – Ben Grossmann Jan 3 '16 at 20:18
• I think I understand what you mean, but doesn't every subsequence of $(a_n)$ converge to 0? – user84413 Jan 3 '16 at 22:55
• Yes, every subsequence of $(a_n)$ converges to $0$. However, your question is about sums of subsequences. – André Nicolas Jan 3 '16 at 22:59