# Sum or upped bound of geometrico-harmonic series

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 P: 2 Hi, I need help to determine the upper bound of this infinite series. $$\sum_{k=p+1}^{\infty} \frac{1}{k} a^k \ \ \ \ ; a \leq 1$$ The paper I am reading reports the upper bound to be, $$\sum_{k=p+1}^{\infty} \frac{1}{k} a^k \leq \frac{1}{p+1}\sum_{k=p+1}^{\infty} a^k = \frac{1}{p+1} \cdot \frac{a^{p+1}}{1-a}$$ I totally cannot understand how could he factor 1/k coefficient out of the summation and replace it with 1/(p+1). Please point me in the right direction.
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 Quote by aamir.ahmed I totally cannot understand how could he factor 1/k coefficient out of the summation and replace it with 1/(p+1). Please point me in the right direction.
k is at least p+1, so 1/k is at most 1/(p+1).

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