21. (SWE 1) Find the greatest integer less than or equal to $\sum_{k=1}^{2^{1983}} k^{1 / 1983-1}$.
21. Using the identity  $$ a^{n}-b^{n}=(a-b) \sum_{m=0}^{n-1} a^{n-m-1} b^{m} $$  with \(a=k^{1 / n}\) and \(b=(k-1)^{1 / n}\) one obtains  $$ 11 \text { and } k \geq 1 $$  This gives us the inequality \(k^{1 / n-1}1\) and \(k \geq 1\). In a similar way one proves that \(n\left((k+1)^{1 / n}-k^{1 / n}\right)1\) and \(k \geq 1\). Hence for \(n>1\) and \(m>1\) it holds that  $$ \begin{aligned} n \sum_{k=1}^{m}\left((k+1)^{1 / n}-k^{1 / n}\right) & <\sum_{k=1}^{m} k^{1 / n-1} \\ & <n \sum_{k=2}^{m}\left(k^{1 / n}-(k-1)^{1 / n}\right)+1, \end{aligned} $$  or equivalently,  $$ n\left((m+1)^{1 / n}-1\right)<\sum_{k=1}^{m} k^{1 / n-1}<n\left(m^{1 / n}-1\right)+1 $$  The choice \(n=1983\) and \(m=2^{1983}\) then gives  $$ 1983<\sum_{k=1}^{2^{1983}} k^{1 / 1983-1}<1984 $$  Therefore the greatest integer less than or equal to the given sum is 1983.