# Doubt with test of condensation of Cauchy

I need to prove test of condensation of Cauchy, i see the proof, but i dont understand some parts of the proof.

If someone can help i'm will very glad

The proof:

Let $a_n$ be nonincreasing and nonnegative (this just follows from $a_{n+1}\leq a_n$) Now we will use the comparison test:

Let $\sum_{n=1}^\infty a_{n}$ be convergent

\begin{align*} a_1+\frac12\sum_{n=1}^K2^na_{2^n}&=a_1+a_2+2a_4+4a_8+\dots+2^{K-1}a_{2^K}\\ &\leq a_1+ a_2+a_3+a_4+a_5+a_6+a_7+a_8+\dots+a_{2^{K-1}+1}+\dots+a_{2^K-1}+a_{2^K}\\ &\leq \sum_{n=1}^\infty a_n \end{align*}

So its partial sums are bounded and $\sum_{n=1}^\infty 2^na_{2^n}$ is convergent.

I dont understand why $2^{K-1}a_{2^{K}}<a_{2^{K-1}+1}+...+a_{2^{K}-1}+a_{2^{K}}$

and why all sum $a_1+ a_2+a_3+a_4+a_5+a_6+a_7+a_8+\dots+a_{2^{K-1}+1}+\dots+a_{2^K-1}+a_{2^K}\\ \leq \sum_{n=1}^\infty a_n$

## 3 Answers

A concrete example might help here. Since $a_n$ is decreasing we have $a_3 > a_4$. Accordingly, $2a_4 = a_4 + a_4 < a_3 + a_4$.

Similarly, we have $a_5 > a_6 > a_7 > a_8$, which implies that $a_8 + a_8 + a_8 + a_8 < a_5+a_6+a_7+a_8$.

This same logic extends to give the general inequality $2^{K-1}a_{2^{K}}<a_{2^{K-1}+1}+...+a_{2^{K}-1}+a_{2^{K}}$

As far as your question, note that all the terms in $a_1+ a_2+a_3+a_4+a_5+a_6+a_7+a_8+\dots+a_{2^{K-1}+1}+\dots+a_{2^K-1}+a_{2^K}$ are terms of $a_n$, but there are only finitely many of them in this sum, while there are infinitely many terms $a_n$. Thus, the sum of all terms $a_n$, i.e. $\sum_{n=1}^\infty a_n$, should be greater (since $a_n$ are non-negative)

Question 1 Why is $2^{K - 1}a_{2^K} \leq a_{2^{K - 1} + 1} + ... a_{2^K}$?

It might be easier to work first with the first few terms; notice that $a_1 + a_2 + a_3 + ... + a_8 \geq a_1 + a_2 + 2a_4 + 4a_8$.

This follows because $a_4 < a_3$, so $2a_4 \leq a_3 + a_4$, and $a_8 < a_5, ..., a_7$, so $4a_8 \leq a_5 + ... + a_8$.

Following this vein of logic, $2^{K - 1}a_{2^K} \leq a_{2^{K - 1} + 1} + ... + a_{2^K}$.

Question 2 This seems trivial to me. $a_1 + a_2 + ... + a_{2^K}$ is the $2^k$th partial sum, so of course it is less than the infinite sum. Feel free to let me know if I'm missing something.

• Actually, the answer below beat me to it; I didn't see it when I was typing! – yandz Apr 2 '17 at 0:29

$2^{K-1}a_{2^{K}}\le a_{2^{K-1}+1}+...+a_{2^{K}-1}+a_{2^{K}}$ because, as the terms $a_n$ are non increasing, each term before $a_{2^{K}}$ is at least equal to $a_{2^{K}}$, in particular $$a_{2^{K}}\le a_{2^{K}-1}, a_{2^{K}-2},\dots, a_{2^{K-1}+1}$$ and there are $2^{K-1}$ of these terms. Just add the corresponding inequalities to obtain what you want.

Concerning the last inequality, it results from the terms being non-negative, the partial sum is at most equal to the total sum.