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I'm having trouble with this particular exercise in limits, and I just can't seem to find a way to crack it.

I saw a similar exercise online where they used integrals, but it's pretty early in the course so we're only supposed to use basic limit arithmetics and the Squeeze theorem (Oh boy, and I thought it had a bad name in MY language). That said, I already tried the Squeeze theorem and it doesn't work.

$$\lim_{n\to\infty} \left( \frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{(3n-2)(3n+1)} \right)$$

What am I missing?

Thanks in advance.

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HINT:

$$\frac1{(3r-2)(3r+1)}=\frac13\frac{3r+1-(3r-2)}{(3r-2)(3r+1)}=\cdots$$

$$\implies3\cdot\frac1{(3r-2)(3r+1)}=\frac1{3r-2}-\frac1{3(r+1)-2}=T(r)-T(r+1)$$ where $T(m)=\dfrac1{3m-2}$

Set a few initial values of $r$ to recognize the Telescoping Series

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  • $\begingroup$ Thanks for replying! In my calculations though, it becomes $\frac{1}{3} \cdot \left( \frac{1}{3n-2} - \frac{1}{3n+1} \right)$. I'll try again and see if I can get a recursive function out of it (or rather find my mistake and get to your answer). Thanks in the meantime. $\endgroup$ – Elad Avron Apr 17 '15 at 15:31

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