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

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The discussion focuses on proving that the infinite sum of 1/(k(k+1)) equals 1 by using the identity 1/k - 1/(k+1) = 1/(k(k+1)). Participants clarify that the sum can be expressed as a telescoping series, where most terms cancel out. The limit of the partial sums as n approaches infinity leads to the conclusion that S_infinity equals 1. There is some confusion regarding the rules of summation and convergence, but the main argument successfully demonstrates the desired result. The thread emphasizes the importance of understanding limits and series in mathematical proofs.
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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?
 
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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...
 
ChelseaL said:
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
 
So basically they are equal?
 
I would write:

$$S_{\infty}=\sum_{k=1}^{\infty}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{\infty}\left(\frac{1}{k}\right)-\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)=1+\sum_{k=2}^{\infty}\left(\frac{1}{k}\right)-\sum_{k=2}^{\infty}\left(\frac{1}{k}\right)$$
 
MarkFL said:
$$S_{\infty}=\sum_{k=1}^{\infty}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{\infty}\left(\frac{1}{k}\right)-\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)=1+\sum_{k=2}^{\infty}\left(\frac{1}{k}\right)-\sum_{k=2}^{\infty}\left(\frac{1}{k}\right)$$
I am not sure I am familiar with the rule that allows writing $$\sum_{k=1}^{\infty}\left(\frac{1}{k(k+1)}\right)=\sum_{k=1}^{\infty}\left(\frac{1}{k}\right)-\sum_{k=1}^{\infty}\left(\frac{1}{k+1}\right)$$.
 
Evgeny.Makarov said:
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:

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