Proof that regular constant-sized deposits always outgrow one-time deposit in the long run?

Question

I want to prove: Given a constant interest rate, no matter how small deposits $\beta$ as long as it goes regularly, we can always outgrow an arbitrarily large one-time lump-sum invested at the start of these constant deposits. Here is my attempt at expressing it mathematically :

Assume $0<\beta<\alpha$ and $\xi >0$

$$\beta \sum_{k=0}^{k=N} (1+\xi)^k > \alpha (1+\xi)^N $$

Can we prove we can always find sufficiently large $N_b$ so that the above is true $\forall N>N_b$ given the constraints on $\xi,\beta,\alpha$ ?

I am interested in both advanced solutions and simpler ones.

Answer 1

Assuming $0<\xi<1$, we have$$ \begin{align}&\beta \sum_{k=0}^{k=N} (1+\xi)^k > \alpha (1+\xi)^N\iff\\ &\beta\frac{(1+\xi)^{N+1}-1}{\xi}>\alpha (1+\xi)^N\iff\\ &\frac{\beta}{\alpha}>\frac{\xi(1+\xi)^N}{(1+\xi)^{N+1}-1}\iff\\ &\frac{\beta}{\alpha}>\frac{\xi}{1+\xi-(1+\xi)^{-N}} \end{align}$$

The last expression goes to $\frac{\xi}{1+\xi}$ as $N\to\infty$ so provided $$\beta>\frac{\alpha\xi}{1+\xi}$$ the periodic deposits will accumulate to a larger amount over a sufficient period of time.

Answer 2

HINT...the left hand side is a geometric series you can sum. Then rearrange the inequality. It looks like this only works if $$\beta>\alpha\xi$$ So $\beta$ cannot be arbitrarily small. But for suitable $\beta$ you can choose $N$ so that$$N>\frac{\log\beta-\log(\beta-\alpha\xi)}{\log(1+\xi)}$$