Asymptotic series

Question

Could somebody explain to me why

$\sum_{n=1}^\infty b^{-n^b}$

is nearly equal to $\frac 1b$ for $b>2$?

Answer 1

For $b>2$,$$S=\frac1b+\frac1{b^{2^b}}+\frac1{b^{3^b}}+\cdots<\frac1b+\frac1{b^4}+\frac1{b^9}+\cdots<\frac1b+\frac1{16}+\frac1{512}+\cdots$$

For $b>3$,$$S=\frac1b+\frac1{b^{2^b}}+\frac1{b^{3^b}}+\cdots<\frac1b+\frac1{b^8}+\frac1{b^{27}}+\cdots<\frac1b+\frac1{256}+\frac1{134217728}+\cdots$$

Below a plot of $\dfrac1S$.

Answer 2

$\newcommand{abs}[1]{\left\lvert{#1}\right\rvert}$As soon as $b\ge2$ $$0\le-\frac1b+\sum_{n=1}^\infty b^{-n^b}=\sum_{n=2}^\infty b^{-n^b}\le\sum_{n=2}^\infty b^{-n^2}\le\sum_{n=2}^\infty b^{-n}=\frac{1}{b^2-b}\to 0$$

With similar estimates (say, the fact that $n^2>kn$ for all $n\ge k$) it is possibile to show that the difference decays faster that any inverse of a polynomial.