An interesting limit

Let \displaystyle s_n=\sum_{k=1}^{\infty} \frac{1}{k(k+1)^n}. Prove that

    \[\lim_{n \rightarrow +\infty}  2^n s_n =1\]

Solution

We note that

    \[s_{n} - s_{n-1} = -\sum_{k=1}^{\infty} \frac{k}{k(k+1)^n}= 1- \zeta(n)\]

Thus,

    \begin{align*} 1 & = 2^n \cdot \frac{1}{2^n} \\ &\leq 2^n \left ( \zeta(n) -1 \right )\\ &=2^n \left ( \frac{1}{2^n} + \sum_{k=3}^{\infty} \frac{1}{k^n} \right ) \\ &\leq 2^n \left ( \frac{1}{2^n} + \int_{2}^{\infty} \frac{{\rm d}x}{x^n} \right ) \\ &= 1+ \frac{2}{n-1} \longrightarrow 1 \end{align*}

and the result follows from the sandwich theorem.

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