How to prove this series $\sum_{j=1}^{\infty} \sqrt{a_j a_{j+1}}$ converges? If $a_j>0$ for every $j$ and if $\sum_{j=1}^\infty a_j$ converges then prove that
$\sum_{j=1}^\infty \sqrt{a_j a_{j+1}}$ converges. I want to use arithmetic mean${}>{}$geometric mean, but I don't know how to show the series $\sum \frac{a_j +a_{j+1}}{2}$ converges.
 A: $0<\sqrt{a_ja_{j+1}}\le \max (a_j,a_{j+1})<\max (a_j,a_{j+1})+\min (a_j,a_{j+1})=a_j+a_{j+1}.$
Let $A(j)=\sum_{n=1}^ja_n.$ Let $b_n=a_n+a_{n+1}.$ Let $B(j)=\sum_{n=1}^jb_j.$
Suppose $A(j)\to L$ as $j\to\infty.$
Given any $\epsilon>0,$ there exists $j_{\epsilon} \in \Bbb N$ such that $\forall j>j_{\epsilon}\,(\,|-L+A(j)|<\epsilon /2 ),$ and hence also $\forall j>j_{\epsilon}\,(\,|-L+A(j+1)|<\epsilon /2).$
Observe that $B(j)=A(j)+A(j+1)-a_1$ and therefore for all $j>j_{\epsilon}$ we have $$|-(2L-a_1)+B(j)|=|(-L+A(j))+(-L+A(j+1)|\le$$ $$\le |L+A(j)|+|-L+A(j+1)| <\epsilon /2+\epsilon /2=\epsilon.$$
So $B(j)$ converges to $2L-a_1$ as $j\to\infty.$
In the 1700's you could say: When $j$ and $j+1$ are infinitely large then $A(j)$ and $A(j+1)$ are infinitely close to $L$ so $B(j)=A(j)+A(j+1)-a_1$ is infinitely close to $L+L-a_1.$
A: \begin{align}
& \sum_{j=1}^\infty \frac{a_j +a_{j+1}}{2} \\[8pt]
= {} & \frac 1 2 \left( \left(\sum_{j=1}^\infty a_j \right) + \left( \sum_{j=1}^\infty a_{j+1} \right) \right) \\[8pt]
= {} & \frac 1 2 \Big( \big( a_1 + a_2 + a_3 + a_4 + \cdots\big) \tag 1 \\
& \qquad\qquad{}+\big(a_2 + a_3 + a_4 + \cdots \big) \Big) \tag 2
\end{align}
The series on line $(1)$ converges if and only if that on line $(2)$ converges since just adding or taking away a single term (although it changes the value of the sum) does not change convergence or divergence.
A: Here's another crack at it: Just observe from the AM-GM inequality: $\sqrt{xy} \le \dfrac{x+y}{2}$ with $x = a_j, y = a_{j+1}$. The use of max and min of two numbers is correct but not necessary for this kind of question.
