## Pseudo-alternate series

Suppose is a non-increasing sequence of positive real numbers and such that is convergent.

Prove that .

The proof for this problem was given by Fedja here. A more detailed and polished version of his answer will be posted below. There are a few interesting ideas, that might be useful in the proof of similar problems, so I’ll divide the proof in a few lemmas.

**Lemma 1.**If the family of sums are bounded by and is an increasing sequence of numbers in , then the sums are bounded by .

*Proof:* The proof is easy if we use Abel’s inequality. Let be a sequence of positive numbers which is decreasing, and be a sequence of real(or complex) numbers. Then the following inequality holds:

where . Using Abel’s inequality we have

where .

Secondly, divide the sequence into blocks of consecutive elements of the sequence such that if is such a block, then . Denote by the index of the first element in the -th block and . Then we can state the next result.

**Lemma 2.** It is enough to prove that .

*Proof:* Let be a positive integer. Then there exists another unique positive integer such that . Then we have

Choose . Because the series is convergent, it follows that the sums can be bounded by for with large enough. Furthermore, we have

and if there exists a such that then Lemma 1 shows us that the terms

are bounded by . For large enough, if we have .

Then, for large enough, we have

Since can be made arbitrarily small, Lemma 2 is now proved.

We are left to deal with the proof of the fact that . First, let’s notice that we have

and by using the same trick as in the proof of Lemma 2, we see that

as . Denote

We know that and . Take in the previous inequality and get

This implies that . Since , we find that and that finishes the proof.

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