<s>[INST] Why is $\left(\sum \limits_{n=1}^{\infty}\frac1{n^3}\right)^3 \gt \sum \limits_{n=1}^{\infty}\frac1{n^4} $? 
$$\displaystyle \left(\sum \limits_{n=1}^{\infty}\frac1{n^3}\right)^3 \gt \sum_{n=1}^{\infty}\frac1{n^4} $$

Why is this true ? 
The LHS can be written either as it is or like this: $\displaystyle \sum \limits_{n=1}^{\infty}\frac1{n^9}$ which is clearly smaller.
 [/INST] In general a power of a sum is different from the sum of powers, $(a+b)^3 = a^3 + b^3 + 3ab(a+b)$ coincides with $a^3 + b^3$ only if either (at least) one of $a,b$ is zero or $a+b = 0$. For sums with more terms or series, conditions for equality aren't so straightforward. Here we have
$$1 < \sum_{n=1}^\infty \frac{1}{n^4} < \sum_{n=1}^\infty \frac{1}{n^3} < \left(\sum_{n=1}^\infty \frac{1}{n^3}\right)^3,$$
the first inequality because the first term in the series is $1$ and all further terms positive, the second because for $n > 1$ we have $\dfrac{1}{n^4} < \dfrac{1}{n^3}$, and the last because $x < x^3$ for all $x > 1$.
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