On the decreasing sequences I do not know how to prove that these sequences are decreasing:
$$
a_n=\frac{1}{\sqrt{n}(n+1)}-\frac{1}{\sqrt{n+1}(n+2)},
$$
$$
b_n=\frac{1}{n(\sqrt{n}+1)}-\frac{1}{(n+1)(\sqrt{n+1}+1)}.
$$
Thank you for all kind help and comments.
My attemption. I considered the following function
$$
f(x)=\frac{1}{\sqrt{x}(x+1)}-\frac{1}{\sqrt{x+1}(x+2)},
$$
and calculated its derivative. But its derivative is rather complicated
$$
\nabla f(x)=\frac{3x+1}{2\sqrt{x}x(x+1)^2}-\frac{3x+4}{2\sqrt{x+1}(x+1)(x+2)^2}
$$
 A: Use the Mean Value Theorem to show there is an $\eta$ between $n$ and $n+1$ so that
$$
\begin{align}
a_n
&=\frac1{\sqrt{n}(n+1)}-\frac1{\sqrt{n+1}(n+2)}\\
&=\frac{\frac32\sqrt{\eta}+\frac12\frac1{\sqrt{\eta}}}{\eta(\eta+1)^2}\\
&=\frac32\frac1{\sqrt{\eta}(\eta+1)^2}+\frac12\frac1{\sqrt{\eta}^3(\eta+1)^2}
\end{align}
$$
which is obviously a decreasing function.
Likewise, show there is an $\eta$ between $n$ and $n+1$ so that
$$
\begin{align}
b_n
&=\frac1{n(\sqrt{n}+1)}-\frac1{(n+1)(\sqrt{n+1}+1)}\\
&=\frac{\frac32\sqrt{\eta}+1}{\eta^2(\sqrt{\eta}+1)^2}\\
&=\frac32\frac1{\eta^{3/2}(\sqrt{\eta}+1)^2}+\frac1{\eta^2(\sqrt{\eta}+1)^2}
\end{align}
$$
which is obviously a decreasing function.
A: Both sequences can be written as $a_n = f(n) - f(n+1)$ where $f$
is a (stricly) convex function.
Then
$$
 a_{n+1} - a_n = \bigl(f(n+1) - f(n+2)\bigr) - \bigl((f(n) - f(n+1)\bigr) = 2f(n+1)- f(n) - f(n+2) < 0
$$
because 
$$
f(n+1) < \frac12 \bigl(f(n) + f(n+2) \bigr)
$$
holds for a strictly convex function.
For your first sequence, $f(x)=\frac{1}{\sqrt{x}(x+1)}$
is convex because 
the second derivative
$$
f''(x) = \frac{5x^2 + 10 x + 3}{4 x^{5/2}(x+1)^3}
$$
is positive for $x > 0$.
For the second sequence, $f(x) = \frac{1}{x (\sqrt{x} + 1)}$
and
$$
f''(x) = \frac{15 x + 21 \sqrt x +8}{4 (x^{3/2} + x)^3} 
$$
is positive for $x > 0$.
(All derivatives calculated with Wolram Alpha :)
A: Observe that for any $n$,
\begin{align}
\sum_{k=1}^n a_k &= \sum_{k=1}^n \left(\frac1{\sqrt k(k+1)} - \frac1{\sqrt{k+1}(k+2)}\right)\\
&= \frac12 - \frac1{\sqrt{n+1}(n+2)}\\
&\stackrel{n\to\infty}\longrightarrow\frac12
\end{align}
and similarly
\begin{align}
\sum_{k=1}^n b_k &= \sum_{k=1}^n \left( \frac1{k(\sqrt k + 1)}\right) - \left( \frac1{(k+1)(\sqrt{k+1} + 1)}\right)\\
&= \frac12 - \left( \frac1{(n+1)(\sqrt{n+1} + 1)}\right)\\
&\stackrel{n\to\infty}\longrightarrow\frac12.\end{align}
Since the sequences of partial sums are monotone increasing and convergent, the sequences $\{a_n\}$ and $\{b_n\}$ are monotone decreasing.
