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This question already has an answer here:

How can you calculate sums such as:

$$\sum_{k=1}^∞{\frac{1}{k(k+1)}}$$

How do you best explain this to students in a rigorous or non-rigorous way?

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marked as duplicate by Travis, 3SAT, Clement C., user147263, Em. Mar 5 '16 at 23:55

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You just do the actual work : by definition, $\sum_{k \ge 1} \frac 1{k(k+1)} = \lim_{N \to \infty} \sum_{k=1}^N \frac 1{k(k+1)}$. Computing the partial sums, you get $$ \sum_{k=1}^N \frac 1{k(k+1)} = \sum_{k=1}^N \frac 1k - \frac 1{k+1} = 1- \frac 1{N+1} $$ because this sum is telescopic (feel free to explicitly write the sum with $\cdots$ dots to your students or expand as a difference of two sums and shift indices). Therefore $\sum_{k \ge 1} \frac 1{k(k+1)} = \lim_{N \to \infty} 1 - \frac 1{N+1} = 1$.

Hope that helps,

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Less than rigorous:

First note that $\dfrac 1 {12\cdot13} = \dfrac 1 {12} - \dfrac 1{13}$ and similarly for other consecutive integers.

So we have $$ \left(1 - \frac 1 2 \right) + \left(\frac 1 2 - \frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) + \cdots. $$ Then $\dfrac{-1}2$ cancels $\dfrac{+1}2$, and $\dfrac{-1}3$ cancels $\dfrac{+1}3$, and so on. Everthing cancels except the first term, which is $1$.

Rigorous: Look at $$ \left(1 - \frac 1 2 \right) + \left(\frac 1 2 - \frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) + \cdots + \left(\frac 1{n} - \frac 1{n+1}. \right) $$ We need $\lim\limits_{n\to\infty}$ of that. Everything cancels except $1 -\dfrac 1{n+1}$. The essential point omitted in the less-than-rigorous version above is that the very last term approaches $0$.

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$$ \sum_{k=1}^∞{\frac{1}{k(k+1)}} = \sum_{k=1}^∞{\frac{1}{k}-\frac{1}{k+1}} = \sum_{k=1}^∞{\frac{1}{k}} - \sum_{k=1}^∞\frac{1}{k+1} = \sum_{k=1}^∞{\frac{1}{k}} - \sum_{k=2}^∞\frac{1}{k} \\\ \\ \begin{align} = 1 &+ 1/2 + 1/3 + 1/4 + ⋯ \\ &- 1/2 - 1/3 - 1/4 - ⋯ \\ \\\ = 1 \end{align} \\\ \\ $$

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    $\begingroup$ You should use the partial sums, not the infinite sums. The quantities you deal with after the first equality do not make sense -- it's $\infty - \infty$. $\endgroup$ – Clement C. Mar 5 '16 at 19:08
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    $\begingroup$ @ClementC. Considering that the OP asked for "a rigorous or non-rigorous way" to compute the sum, I think this answer is pretty good. $\endgroup$ – zz20s Mar 5 '16 at 19:13
  • $\begingroup$ @zz20s I really don't see the point of doing this the "non-rigorous" way, since it is exactly the same proof as with the partial sums... but without correctness. What is the benefit then? $\endgroup$ – Clement C. Mar 5 '16 at 19:15
  • $\begingroup$ I agree with you, but that's something to take up with the OP. $\endgroup$ – zz20s Mar 5 '16 at 19:16
  • $\begingroup$ @ClementC.: For the sake of explaining it to people far less experienced at mathematics than you. Sometimes I feel like people on this site have studied mathematics for so long that they've forgotten how confusing something like partial sums can be to a beginner. $\endgroup$ – Zaz Mar 5 '16 at 19:26
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While exploitation of the resulting telescoping series after partial fraction expansion is a very simple way forward, I thought it might be instructive to present another way forward. Here, we write

$$\begin{align} \sum_{k=1}^\infty \frac{1}{k(k+1)}&=\sum_{k=1}^\infty \frac1k \int_0^1 x^k\,dx\\\\ &=\int_0^1\sum_{k=1}^\infty \frac{x^k}{k}\,dx\\\\ &=-\int_0^1\log(1-x)\,dx\\\\ &=1 \end{align}$$

as expected!

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