MHB Proving \sum_{r=1}^{\infty} \frac{1}{k(k+1)} = 1 Using Deduction

[ChelseaL]

Use the fact that \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)} to show that


Deduce that

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

How do I solve this?

 

[MarkFL]

Take a look at what I posted for the $n$th sum of this same series, and see if you can't change in over to an infinite sum...

 

[topsquark]

Use the fact that \frac{1}{k} - \frac{1}{k+1} = \frac{1}{k(k+1)} to show that


Deduce that

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

How do I solve this?

Are you summing over r or over k? I'll assume k.

[math]\sum _{k = 1}^{\infty} \frac{1}{k(k + 1)} = \sum _{k = 1} ^{\infty} \left ( \frac{1}{k} - \frac{1}{k + 1} \right )[/math]. You can break this up into two summations and evaluate them with the known rule, or you could note that this is a telescopic series and most of the terms will cancel out.

-Dan

 

[ChelseaL]

So basically they are equal?

 

[MarkFL]

I would write:

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

 

[Evgeny.Makarov]

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

I am not sure I am familiar with the rule that allows writing $$\sum_{k=1}^{\infty}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{\infty}\left(\frac{1}{k}\right)-\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)$$.

 

[MarkFL]

I am not sure I am familiar with the rule that allows writing $$\sum_{k=1}^{\infty}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{\infty}\left(\frac{1}{k}\right)-\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)$$.

Yes, that was a bad post. The OP stated an unfamiliarity with limits, so I "winged it." What I would actually do is:

$$S_{\infty}=\lim_{n\to\infty}\left(\sum_{k=1}^{n}\left(\frac{1}{k(k+1)}\right)\right)=\lim_{n\to\infty}\left(\frac{n}{n+1}\right)=1$$

 

[topsquark]

I am not sure I am familiar with the rule that allows writing $$\sum_{k=1}^{\infty}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{\infty}\left(\frac{1}{k}\right)-\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)$$.

Ah, convergence issues. (Swearing) One of my favorite topics (to ignore.)

Thanks for the catch.

-Dan