<s>[INST] How to find the upper bound for this series I came across this series in a book and the author tells the upper bound and it checks out under the conditions but i could n't find the connection or the reason if u may, that had i not been presented with this info, that i could have used to find the upper bound myself.
For $s>1$ the sum of $2^{p}-1$ terms of the series is less than 
$\dfrac {1} {1^{s-1}}+\dfrac {1} {2^{s-1}}+\dfrac {1} {4^{s-1}}+\dfrac {1} {8^{s-1}}+\ldots +\dfrac {1} {2^{\left( p-1\right) \left( s-1\right) }} < \dfrac {1} {1-2^{1-s}}$
Any hints or clues would be much appreciated.
 [/INST] Use the fact that if $|x|<1$, then $\frac{1}{1-x}$ has the power series expansion
$$\frac{1}{1-x}=1+x+x^2+x^3+\cdots +x^n+\cdots .$$
Or, in less fancy language, if $|x|<1$, then the infinite  geometric series $1+x+x^2+\cdots+x^n+\cdots$ has sum $\frac{1}{1-x}$. In your particular example, $x=\frac{1}{2^{s-1}}$. Since $s>1$, we have $|x|<1$. 
More explicitly, for this value of $x$, the left-hand side is just
$$1+x+x^2+\cdots+x^{p-1}.$$
This finite geometric series has sum 
$$1+x+x^2+\cdots+x^{p-1}=\frac{1-x^p}{1-x}=\frac{1}{1-x}-\frac{x^p}{1-x}.$$
This says that the sum on the left-hand side is less than $\frac{1}{1-x}$.  It even says by how much it is less, namely $\frac{x^p}{1-x}$.
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