Computing the sum of the ratios: \(\displaystyle{S}_{{n}}={\sum_{{{k}={0}}}^{{n}}}{\frac{{{n}-{k}+{1}}}{{{\left({n}+{1}\right)}{\left({n}+{2}\right)}{\left({n}-{k}+{2}\right)}}}}\)

tibukooinm

tibukooinm

Answered question

2022-03-27

Computing the sum of the ratios:
Sn=k=0nnk+1(n+1)(n+2)(nk+2)

Answer & Explanation

Leonardo Mcpherson

Leonardo Mcpherson

Beginner2022-03-28Added 13 answers

You can write the sum as
1(n+1)(n+2)k=0nn+1kn+2k
so it's enough to compute the last part of it. This is just
n+1n+2+nn+1++n+1nn+2n=11n+2+11n+1++11n+2n
=n+2Hn+2
where Hn=k=1n1k is the n:th Harmonic number. Your sum is then
n+2Hn+2(n+1)(n+2)

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