<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] For $n>1$ we have $$1/n^5+1/(n+1)^5+1/(n+2)^5+...<$$ $$<\frac {1}{n^3}\left(\frac {1}{(n-1)n}+\frac {1}{n(n+1)}+\frac {1}{(n+1)(n+2)}+...\right)=$$ $$=\frac {1}{n^3}\left((\frac {1}{n-1}-\frac {1}{n})+(\frac {1}{n}-\frac {1}{n+1})+(\frac {1}{n+1}-\frac {1}{n+2})+...\right)=$$ $$=\frac {1}{n^3(n-1)}$$ because the sum of the first $k$ terms of the last infinite series above is $\frac {1}{n-1}-\frac {1}{n-1+k}.$
This is not as sharp an upper bound as can be found by deeper methods. We have $1.036<\sum_{n=1}^{n=5}n^{-5}<1.037$ and the sum of the remaining terms is less than $1/(6^3\cdot 5)=1/1080.$ So the sum rounded to 2 decimal places is $1.04$
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