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I have this series:

$${1\over 1\cdot3}+{1\over 3\cdot5}+{1\over 5\cdot7}+. . .$$

I am supposed to choose a test that will prove that this series converges. I think that it looks like a telescoping series, but I have no idea how to find the general term for it. I haven't been able to find much help online. I can see that this series probably converges to zero, but that's not an appropriate answer to this problem.

How do I find the general term?

Edit: The other question does not address how this was found:

$${1\over (2n-1)(2n+1)}$$

I don't know how to find that on my own, and that is what I'm looking for.

matryoshka
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3 Answers3

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The general term $$t_r=\frac{1}{(2r-1)(2r+1)}=\frac{1}{2}[\frac{(2r+1)-(2r-1)}{(2r+1)(2r-1)}]=\frac{1}{2}[\frac{1}{2r-1}-\frac{1}{2r+1}]$$

The partial sum $$S_n = \sum_{r=1}^n t_r =\frac{1}{2}[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{2n-1}-\frac{1}{2n+1}]=\frac{1}{2}[1-\frac{1}{2n+1}]$$ $$=\frac{n}{2n+1}$$

and hence $$S=\lim_{n\to\infty} S_n =\lim_{n\to\infty} \left(\frac{1}{2+\frac{1}{n}}\right)=\frac{1}{2}$$

SchrodingersCat
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  • How did you find ${1\over (2r-1)(2r+1)}$ ? That's the step I'm having trouble with. The rest I'm fine with – matryoshka Oct 27 '15 at 23:51
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    This thing is something which has no specific algorithm. You need to practise more and more problems and only then you will be able to master that ability. In general, you need to observe the way the sequence has different values for different $n$. For example in this case, observe the denominator. You will find them as products of 2 odd numbers. Now odd numbers are of the form $2n\pm1$. Use that and then check how you can get the general term. – SchrodingersCat Oct 28 '15 at 13:30
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Notice.,

General $n$th term of the given series $$T_n=\frac{1}{(2n-1)(2n+1)}=\frac{1}{2}\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)$$ Now, the sum of the given series

$$S_n=\sum_{n=1}^{\infty} T_n$$ $$=\lim_{n\to \infty}\frac{1}{2}\left(\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{5}{1}-\frac{1}{7}\right)+\ldots +\left(\frac{1}{2n-1}-\frac{1}{2n+1}\right)\right)$$

$$=\lim_{n\to \infty}\frac{1}{2}\left(1-\frac{1}{2n+1}\right)$$ $$=\frac{1}{2}\left(1-0\right)=\color{red}{\frac{1}{2}}$$

Harish Chandra Rajpoot
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The general term of the corresponding sequence is $$\frac1{(2n-1)(2n+1)}=\frac12\left(\frac1{2n-1}-\frac1{2n+1}\right)$$

So the general term of the series is $$\frac12-\frac1{2(2n+1)}$$

Now can you take the limit?

Macavity
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