Summation of series (proof)  
I need help with this question. I tried, but couldn't do it. How do I approach it? 
 
I tried this approach. The problem is, even though it's obvious (by intuition) that the series (A1+A2+A3+......) adds up to less than 1, I can't show it that it does indeed add up to less than 1. Because the series is irregular, without a pattern. How do I prove it? I don't even know if this is the right approach, so please help me with it. Thanks
 A: First you can prove that the sum converges (this step is not necessary, but interesting by itself) :
$$\frac{A_n}{A_{n-1}} = 1-\frac{3}{2n}$$
Find a new sequence of the form $B_n=n^{-\alpha}$ such that $\alpha>1$ and $\frac{B_n}{B_{n-1}}>\frac{A_n}{A_{n-1}}$ :
$$\frac{B_n}{B_{n-1}} = (\frac{n-1}{n})^\alpha = 1-\frac{\alpha}{n}+o(\frac1n)$$
If you take $\alpha=\frac54$, you can conclude that $\frac{B_n}{B_{n-1}}>\frac{A_n}{A_{n-1}}$ for sufficiently large $n$, which leads to $A_n<B_n$ for $n>N_0$. As $\sum B_n$ converges, so does $\sum A_n$ (this is known as the Raabe-Duhamel criterion in France).
Now note $S_n=\sum_{k=1}^n A_k$. From $2nA_n=(2n-3)A_{n-1}$ follows :
$$\sum_{k=1}^n 2kA_k = 2A_1+\sum_{k=2}^n (2k-3)A_{k-1} = 1+\sum_{k=1}^{n-1} (2k-1)A_k = 1 + \sum_{k=1}^{n-1} 2kA_k - S_{n-1}$$
By subtracting the terms present on both side of this last equation, you are down to :
$$S_{n-1}=1-2nA_n$$
This by itself is enough to prove $S_n<1$ for all $n$ (the convergence of $\sum A_n$ is useless, it is sufficient to prove that $A_n>0$ for all $n$), but it gives the sum of the series :
$$\sum_{k=1}^\infty A_n=1$$
This is because we proved that $A_n<n^{-5/4}$ for sufficiently large $n$.
