How to compute an upper bound of $1 + \frac{1}{2^a} + \frac{1}{3^a} + \cdots$?

Question

Suppose $a>2$.

How to compute an upper bound of $1 + \frac{1}{2^a} + \frac{1}{3^a} + \cdots$ ?

Is $\frac{1}{a-2}$ an upper bound?

Answer 1

The expected bound does not work for large values of $a$.

$\frac a {a-1}$ is an upper bound. Note that $\int_k^{k+1} \frac 1 {x^{a}} dx >\frac 1 {(k+1)^{a}}$. Sum this over $k$ from $1$ to $\infty$ and then add $1$ to get a proof.

[The question is more interesting when $a$ is close to $2$ and there we have $\frac a {a-1} <\frac 1 {a-2}$].

Answer 2

When a = 2 we can say

$1 + \frac 14 (1 + \frac {4}{9}) + \frac {1}{16}( 1 + \frac {16}{25} + \frac {16}{36} + \frac {16}{49}) + \cdots < 1 + \frac 14 (2) + \frac 1{16} (4) + \cdots = 2$

when $a> 1,$

$\sum_\limits{n=0}^{\infty} \frac {1}{n^a} < \sum_\limits{n=0}^{\infty} (\frac {1}{2^{a-1}})^n = \frac {2^{a-1}}{2^{a-1}-1}$