MHB Improper integral from 1 to infinity

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The discussion centers on solving an improper integral from 1 to infinity, specifically focusing on a step involving logarithmic manipulation. The user expresses confusion about transitioning from a logarithmic expression to a simplified form. Another participant clarifies the algebraic steps, showing how to derive the expression (1 - 1/(b+1)) from the original equation. The conversation highlights the importance of understanding logarithmic properties and algebraic simplification in calculus problems. Overall, the exchange emphasizes collaborative problem-solving in mathematical contexts.
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Hello everyone,

I am stuck on this homework problem. I got up to (ln (b / (b+1) - ln 1 / (1+1) ) but I'm not sure how to go to the red boxed step where they have (1 - 1 / (b+1) )

if anyone can figure it out Id really appreciate it.

thank you very much.
 

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$\frac{b}{b+1} = \frac{b+1-1}{b+1} = \frac{b+1}{b+1} + \frac{-1}{b+1} = 1 - \frac{1}{b+1}$
 
THANK YOU soo much for your help and time. Really appreciate it!
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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