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

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 $n$.
