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

[ChelseaL]

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?

 

[MarkFL]

What they want you to do is:

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

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

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

What are you left with?

 

[ChelseaL]

S_n = 1 + (1/n+1)?

 

[MarkFL]

S_n = 1 + (1/n+1)?

To properly use bracketing, you want:

S_n = 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:

\(\displaystyle S_n=1-\frac{1}{n+1}\)

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

\(\displaystyle S_n=\frac{n}{n+1}\)

 

[ChelseaL]

Thank you!