Problem - 3174
Let $\{a_n\}$ be an increasing geometric sequence satisfying $a_1+a_2=6$ and $a_3+a_4=24$. Let $\{b_n\}$ be another sequence satisfying $b_n=\frac{a_n}{(a_n-1)^2}$. If $T_n$ is the sum of first $n$ terms in $\{b_n\}$, show that for any positive integer $n$, it always holds that $T_n < 3$.
Let $r$ be the common ratio of $\{a_n\}$. Then $$\frac{a_3 + a_4}{a_1+a_3}=\frac{a_1r^2+a_2r^2}{a_1+a_2}\implies r^2=\frac{24}{6}=4 \implies r = 2\quad (\because r > 1)$$
It follows that $$a_1+a_2=a_1(1+r)=3a_1 \implies 3a_1=6\implies a_1 = 2$$
$$\therefore\quad a_n = a_1r^{n-1}=2\times 2^{n-1}=2^n$$
Hence, $$b_n=\frac{a_n}{(a_n-1)^2}=\frac{2^n}{(2^n-1)^2} < \frac{2^n}{(2^n-1)(2^n-2)}=\frac{2^{n-1}}{(2^n-1)(2^{n-1}-1)}$$
It is easy to verify that $$\frac{2^{n-1}}{(2^{n-1}-1)(2^n-1)}=\frac{1}{2^{n-1}-1}-\frac{1}{2^n-1}$$
Therefore
\begin{align*}
T_n = & b_1 + b_2 + \cdots + b_n \\
= & 2 + \Big(1-\frac{1}{3}\Big) + \Big(\frac{1}{3}-\frac{1}{7}\Big) + \cdots + \Big(\frac{1}{2^{n-1}-1}-\frac{1}{2^n-1}\Big)\\
= & 3 - \frac{1}{2^n-1} < 3
\end{align*}