I am given the $n$th term of sequence $$T_n = \frac{1}{(4n+1)(4n+2)(4n+3)}.$$ I am struggling in computing the sum $\sum_{n=1}^{\infty} T_n$.
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2Please use MathJax. Here is a tutorial. – Dietrich Burde Apr 16 '24 at 21:32
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See https://artofproblemsolving.com/community/c6h1708880p11011692 – Robert Israel Apr 16 '24 at 21:48
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2Try decomposition to some kind of fraction. – hamam_Abdallah Apr 16 '24 at 21:50
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1Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Apr 16 '24 at 22:07
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Not quite the same as https://math.stackexchange.com/questions/735794/series-for-envelope-of-triangle-area-bisectors as that was $\sum\limits_{n=1}^{\infty}\dfrac{1}{(4n-1)(4n)(4n+1)} = \frac{3}{4} \log_e(2) - \frac{1}{2}$ – Henry Apr 16 '24 at 22:38
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Does this answer your question? Calculating $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{5\cdot 6\cdot 7}+\frac{1}{9\cdot 10\cdot 11}+\cdots$ – peterwhy Apr 16 '24 at 23:07
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The answer is in https://oeis.org/A001505 : $\sum_{n=0}^\infty \frac{1}{(4n+1)(4n+2)(4n+3)} = \frac14 \log 2$, and you need to subtract the term with $n=0$, which is 1/6.
R. J. Mathar
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