MHB Solve Sigma 1/(k(k+1)): Step-by-Step Guide

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The discussion focuses on deriving the formula for the sum S_n = Σ(1/(k(k+1))) from k=1 to n. It begins by using the identity 1/k - 1/(k+1) to express the sum in terms of simpler fractions. The participants clarify the steps to isolate terms and correctly simplify the expression, ultimately arriving at S_n = 1 - 1/(n+1). A suggestion is made to use LaTeX for better readability of mathematical expressions. The final result is confirmed as S_n = n/(n+1), completing the derivation.
ChelseaL
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Use the fact that \frac{1}{k} -\frac{1}{k+1} = \frac{1}{k(k+1)} to show that

n
sigma (\frac{1}{k(k+1)}) = 1-\frac{1}{n+1}
r=1

What do I need to do to solve it?
 
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What they want you to do is:

$$S_n=\sum_{k=1}^n\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^n\left(\frac{1}{k}\right)-\sum_{k=1}^n\left(\frac{1}{k+1}\right)=\sum_{k=1}^n\left(\frac{1}{k}\right)-\sum_{k=2}^{n+1}\left(\frac{1}{k}\right)$$

Now, take off the first term of the first sum, and the last term of the second sum like so:

$$S_n=1+\sum_{k=2}^n\left(\frac{1}{k}\right)-\sum_{k=2}^{n}\left(\frac{1}{k}\right)-\frac{1}{n+1}$$

What are you left with?
 
Sn = 1 + (1/n+1)?
 
ChelseaL said:
Sn = 1 + (1/n+1)?

To properly use bracketing, you want:

Sn = 1 + 1/(n+1)

I would highly recommend learning to use $\LaTeX$ to make your expressions more readable. :)

However, that is incorrect, as you've got the wrong sign in front of the second term. What you want is then:

$$S_n=1-\frac{1}{n+1}$$

This is what we've been asked to show. I would choose to write it this way though:

$$S_n=\frac{n}{n+1}$$
 
Thank you!
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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