Proof an Inequality of a sequence $\frac{1}{2^1-1}+\frac{1}{2^2-1}+\cdots+\frac{1}{2^n-1}\lt \frac{5}{3}$ I have tried to amplify it like this 
 $$\frac{1}{2^1-1}+\frac{1}{2^2-1}+\cdots+\frac{1}{2^n-1}\lt 1+\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2^{n-1}}=2-\frac{1}{2^{n-1}} $$
which obviously too big. Any proper ways to amplify it ?
 A: Let $$ T = \sum_{n=1}^{\infty} \frac{1}{2^n-1}.$$
$$T < \frac{1}{1} + \frac{1}{3} + \sum_{n=3}^{\infty} \frac{1}{2^n-2} = \frac{4}{3} + (\frac{T}{2} - \frac{1}{2}). $$
Hence $ T < \frac{5}{3}$.

Based on what you were trying to do, I'm guessing you wanted to use induction by strengthening the induction hypothesis. If so, this is how you can proceed to solve the problem in a logical manner.
The goal is to prove a statement of the form
$$\sum_{i=1}^n \frac{1}{2^i-1} < \frac{5}{3} - f(n)$$
In order to apply induction easily, the induction step requires $\frac{1}{2^{n+1}-1} < f(n) - f(n+1) $.
Letting $f(n) = \frac{1}{2^n-2}$ satisfies that constraint. (You can determine this form by testing a few fractions. $\frac{1}{2^n}$ doesn't work.)  
It remains to find a base case for which the statement is true. Note that it is not true for $n = 1$. 
Note: The final 2 paragraphs motivate/explain the 2 "magic steps" in the first solution.   
A: Maybe you are overdoing it. Consider leaving a first few terms untouched and then "amplify" the rest like you did. 
$$S < \dfrac{1}{2-1}+\dfrac{1}{2^2-1}+\dfrac{1}{2^3}+...+\dfrac{1}{2^n}=...?$$
