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Sum combined with product


\displaystyle \mathcal{S}= \sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{1+k \log k}{2 + (k+1)\log(k+1)}

(Cornel Ioan Valean)

Solution [by Demetres Christofides]

It is easy to prove inductively that

    \[\sum_{n=1}^{m} \prod_{k=1}^{n} \frac{a_k}{1+a_k} = a_1 - \frac{a_1 \cdot a_2 \cdots a_m}{\left ( 1+a_1 \right )\left ( 1+a_2 \right )\cdots \left ( 1+a_m \right )}\]

Taking a_k= 1+k \log k we will see that the desired sum is equal to 1. Since

    \[\frac{a_k}{1+a_k} = 1 - \frac{1}{1+a_k} \leq e^{1/(1+a_k)}\]

we have that

    \[\prod_{k=2}^{N} \frac{a_k}{1+a_k} \leq \exp \left ( -\sum_{k=2}^{N} \frac{1}{1+a_k} \right ) = \exp \left ( - \sum_{k=2}^{N} \frac{1}{2+ k \log k } \right )\]

and since the series \displaystyle \sum_{k=2}^{\infty} \frac{1}{2+ k \log k } divirges we immediately see that

    \[\prod_{k=2}^{\infty}\frac{a_k}{1+a_k} =0\]

and hence \mathcal{S}=1.

The exercise can also be found at mathematica.gr .

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