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What is value of $\dfrac {1}{2\cdot 3} + \dfrac {1}{4\cdot 5} +\dfrac {1}{6\cdot7} + \cdots =?$

1) $\log \left( \dfrac 2e \right)$

2) $\log \left( \dfrac e2 \right)$

3) $\log \left( 2e \right)$

4) $e-1$

I know it converges but I am not getting how to solve

  • 2
    $$\frac 1{n(n+1)}=\frac 1n-\frac 1{n+1}$$ have you heard the term "telescopic"? – 2017-02-01
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    In $2.3$, is the $.$ a decimal point or multiplying? – 2017-02-01
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    @S.C.B. it's multiplying – 2017-02-01
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    @LOL I answered as if it was a decimal point. Make that a bit clear later on. – 2017-02-01
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    @polfosol no,never heard,I will be very thankful if somebody answer without using strange terms – 2017-02-01
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    @LOL Where did I use strange terms? – 2017-02-01
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    @S.C.B. never mind, but I didn't understand anything – 2017-02-01
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    @LOL Which part didn't you understand? – 2017-02-01
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    @MartinR hey,there is no **limit** in my question,it can't be _duplicate_ – 2017-02-01
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    @LOL Martin R deleted his comment. – 2017-02-01
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    @LOL Also, a series is a limit. – 2017-02-01
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52841/discussion-between-lol-and-s-c-b). – 2017-02-01
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    @LOL.: So what does $+ \ldots$ mean if not a limit? But that other question was different, that was an error on my part. – 2017-02-01
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    Possible duplicate of http://math.stackexchange.com/questions/636496/evaluation-of-series. – 2017-02-01
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    @MartinR you looks so active to mark questions **duplicate** you can see that [here](http://math.stackexchange.com/questions/636496/evaluation-of-series) (s)he is trying by **integration** – 2017-02-01
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    @LOL.: The other poster tried something (which did not work), and you did not show anything that you tried, but it is still a question about the same series. And if you look at the answer to the other question, that is essentially the same as the answer posted below. – 2017-02-01
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    Just to avoid confusion, I did not copy lab's answer. – 2017-02-01
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    @S.C.B.: I did not claim that, and I apologize if I made that impression. – But (as I see it) it is a duplicate question. – 2017-02-01
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    @MartinR I know. So I didn't say "As Martin R seems to be saying...". I said just to avoid confusion, as I would hate for someone to misunderstand that. I did not mean that you either explicitly or implicitly claimed that. Yes, this does seem to be a duplicate according to the policies of this site. – 2017-02-01

3 Answers 3

4

We use $$\frac{1}{n(n+1)}=\frac{n+1-n}{n(n+1)}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$$ Note that your series, $$\dfrac {1}{2 \times 3} + \dfrac {1}{4 \times 5} +\dfrac {1}{6 \times 7} + \cdots=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\dots$$ Since $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots=\ln 2$$ This result is known, and is discussed as the Alternating Harmonic Series on Wikipedia. So since $$\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\dots=1-\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots \right)$$ the answer is $1-\ln 2$. So the answer to your question is $$1-\ln 2=\ln e-\ln 2=\ln \frac{e}{2}$$So the answer is $2)$

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    But answer needs to be in terms of "e" and "$\log$" – 2017-02-01
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    @LOL, see my edit. – 2017-02-01
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    There is no "e" :( – 2017-02-01
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    @LOL $\ln x =\log_{e}x$. – 2017-02-01
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    @LOL. instead of complaining about the formatting of an answer, I recommend you first try to understand the answer, and only afterwards judge whether or not it is effective. – 2017-02-01
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    Check my edit . – 2017-02-01
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    @LOL Is $\log$ the $\log_{10}$ or $\log_{e}$? – 2017-02-01
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    May be $\log_{10}$ otherwise it would be $\text{In}$ – 2017-02-01
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    @LOL I answered and editted. – 2017-02-01
2

Herein we present a methodology that relies on only straightforward arithmetic and knowledge of Riemann sums. To that end, we now proceed.

The general term in the series is $a_n=\frac{1}{2n(2n+1)}=\frac{1}{2n}-\frac{1}{2n+1}$.

Then, we have

$$\begin{align} \sum_{n=1}^N a_n&=\sum_{n=1}^{N}\left(\frac{1}{2n}-\frac{1}{2n+1}\right)\\\\ &=-\sum_{n=1}^{N}\left(\frac{1}{2n}+\frac{1}{2n+1}\right)+\sum_{n=1}^N\frac1n\\\\ &=1-\left(\sum_{n=1}^{2N+1}\frac1n-\sum_{n=1}^{N}\frac1n\right)\\\\ &=1-\sum_{n=N+1}^{2N+1}\frac1n\\\\ &=1-\sum_{n=1}^{N+1}\frac1{n+N}\\\\ &=1-\frac1N\sum_{n=1}^{N+1}\frac{1}{1+n/N}\tag 1 \end{align}$$

Recognizing the sum on the right-hand side of $(1)$ as the Riemann sum for $\int_0^1 \frac{1}{1+x}\,dx=\log(2)$, we have

$$\lim_{N\to \infty}\sum_{n=1}^Na_n=1-\log(2)=\log(e/2)$$

1

An alternative solution, just for fun: $$\begin{eqnarray*}\sum_{n\geq 1}\frac{1}{2n(2n+1)} = \int_{0}^{1}\sum_{n\geq 1}\frac{x^{2n}}{2n}\,dx &=& -\frac{1}{2}\int_{0}^{1}\log(1-x^2)\,dx \\(IBP)\quad&=&\int_{0}^{1}x\cdot\frac{1-x}{1-x^2}\,dx\\&=&\int_{0}^{1}\frac{x}{1+x}\,dx\\&=&1-\log 2=\color{red}{\log\left(\frac{e}{2}\right)}.\end{eqnarray*}$$