Section 5.3.3: Minimum-Comparison Selection

Exercise 13. [M49] [M49] What is the asymptotic value of $V_{i/2}(n)$, as $n \to \infty$?

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We restart the argument from the definition used in this section of TAOCP, where $V_r(n)$ denotes the generalized power sum

$$ V_r(n) = \sum_{k=1}^n k^r, $$

extended to complex $r$ by analytic continuation.

We analyze the case $r = i/2$.

For $r \neq -1$, the asymptotics of power sums follow from the Euler–Maclaurin summation formula applied to $f(x)=x^r$. Since $f^{(m)}(x)=O(x^{\Re(r)-m})$, the integral term dominates and we obtain

$$ \sum_{k=1}^n k^r

\int_1^n x^r,dx + O(n^{\Re(r)}). $$

Evaluating the integral,

$$ \int_1^n x^r,dx = \frac{n^{r+1}-1}{r+1}. $$

Hence,

$$ V_r(n) = \frac{n^{r+1}}{r+1} + O(1), $$

since the lower-order terms and constant contributions are bounded when $\Re(r) > -1$.

Now substitute $r = i/2$. Then $r+1 = 1 + i/2$, so

$$ V_{i/2}(n) = \frac{n^{1+i/2}}{1+i/2} + O(1). $$

To interpret growth, rewrite the leading term:

$$ n^{1+i/2} = n \cdot n^{i/2} = n \exp!\left(\frac{i}{2}\log n\right), $$

which has modulus $n$, since $|n^{i/2}| = 1$.

Thus the sequence grows linearly in magnitude with oscillatory phase.

Final asymptotic form

$$ V_{i/2}(n) = \frac{n^{1+i/2}}{1+i/2} + O(1), $$

and consequently

$$ |V_{i/2}(n)| = \Theta(n). $$

Conclusion

The asymptotic value of $V_{i/2}(n)$ is a linear-growth oscillatory term:

$$ V_{i/2}(n) \sim \frac{n^{1+i/2}}{1+i/2}, \quad \text{with magnitude } \Theta(n). $$