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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?

Zaz
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4 Answers4

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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,

Patrick Da Silva
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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 $0$.

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

Zaz
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    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 $\infty - \infty$. – Clement C. Mar 05 '16 at 19:08
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    @ClementC. Considering that the OP asked for "a rigorous or non-rigorous way" to compute the sum, I think this answer is pretty good. – zz20s Mar 05 '16 at 19:13
  • @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? – Clement C. Mar 05 '16 at 19:15
  • I agree with you, but that's something to take up with the OP. – zz20s Mar 05 '16 at 19:16
  • @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. – Zaz Mar 05 '16 at 19:26
  • @Zaz It may be an subjective issue, but I'd deem manipulations on indices of infinite sums much more confusing to anyone. For the partial sums, you can actually write down the terms on a board, one below the other, and cross them to show they cancel; and then taking the limit of $1-\frac{1}{N+1}$ is quite simple to grasp. $$ \begin{align} &\frac{1}{1}-\frac{1}{2} \
    • &\frac{1}{2}-\frac{1}{3} \
    • &\frac{1}{3}-\frac{1}{4} \

    &\vdots\

    • &\frac{1}{N-1}-\frac{1}{N} \
    • &\frac{1}{N}-\frac{1}{N+1}

    \end{align} $$

     – Clement C. Mar 05 '16 at 19:29
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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!

Mark Viola
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