<s>[INST] Evaluate $\sum_{n=1}^{\infty}{1\over n^5}$ up to the second decimal place I am trying to evaluate
$$\sum_{n=1}^{\infty}{1\over n^5}$$
up to the second decimal place. While the series is convergent, I have no idea how to construct such a bound, preferably using basic properties of series and sequences. Any hints?
 [/INST] Let $N$ be such that $\sum \limits _{k \ge 2^N} ^\infty \frac 1 {n^5} \le \epsilon$. Notice that
$$\sum _{k \ge 2^N} ^\infty \frac 1 {n^5} \le \sum _{m \ge 0} \ \sum _{k = 2^{N+m}} ^{2^{N+m+1}-1} \frac 1 {n^5} \le \sum _{m \ge 0} \ \sum _{k = 2^{N+m}} ^{2^{N+m+1}-1} \frac 1 {(2^{N+m})^5} = \sum _{m \ge 0} \frac {2^{N+m}} {(2^{N+m})^5} = \sum _{m \ge 0} \frac 1 {(2^{N+m})^4} = \\
\frac 1 {2^{4N}} \sum _{m \ge 0} \frac 1 {2^{4m}} = \frac 1 {2^{4N}} \frac 1 {1 - \frac 1 {16}} = \frac 1 {15} \frac 1 {16^{N-1}} .$$
It follows that in order to have the inequality that we have begun with, it is sufficient to impose that $\frac 1 {15} \frac 1 {16^{N-1}} \le \epsilon$, which means $16^{N-1} \ge \frac 1 {15 \epsilon}$, whence $N-1 \ge \log_{16} \frac 1 {15 \epsilon} = -\log_{16} (15\epsilon) = -\frac 1 4 \log_2 (15 \epsilon)$, so $N = \left[ 1 -\frac 1 4 \log_2 (15 \epsilon) \right] + 1$, where $[x]$ denotes the integer part of $x$ (the "floor" function, as some call it).
Now just choose $\epsilon = 10^{-3}$ (because you want the first 2 decimal places to be exact) in order to get $N = \left[ 1 -\frac 1 4 \log_2 (15 \cdot 10^{-3}) \right] + 1 = 3$. This means that
$$\sum _{n=1} ^{2^3-1} \frac 1 {n^5} = 1.\color{red}{03}6849\dots$$
has the first 2 decimal places correct.
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