Using Cauchy's integral test, examine the convergence
Using Cauchy's integral test, examine the convergence of: $$ \sum_{n=1}^{\infty} \frac{1}{n^\alpha}. $$
Explanation of Method
For $\alpha>0$, $f(x)=x^{-\alpha}$ is positive, continuous, and decreasing on $[1,\infty)$, so the integral test applies directly. Evaluating $\int_1^\infty x^{-\alpha}\,dx$ splits into three cases depending on whether $1-\alpha$ is negative, zero, or positive, yielding the classical $p$-series threshold at $\alpha=1$.
- Verify $f(x)=x^{-\alpha}$ satisfies the integral test's hypotheses for $\alpha>0$.
- Evaluate $\int_1^\infty x^{-\alpha}\,dx$ for $\alpha\ne1$ and $\alpha=1$ separately.
- Apply the integral test, including the degenerate case $\alpha\le0$.
Step 1 · Verify the hypotheses
For $\alpha>0$: $f(x)=x^{-\alpha}>0$ and continuous on $[1,\infty)$; $f'(x)=-\alpha x^{-\alpha-1}<0$, so $f$ is decreasing. (For $\alpha\le0$, the terms $1/n^\alpha$ don't tend to $0$, so the series diverges immediately by the divergence test.)
Step 2 · Evaluate the integral
For $\alpha\ne1$:
$$ \int_1^{\infty} x^{-\alpha}\,dx = \lim_{t\to\infty}\frac{t^{1-\alpha}-1}{1-\alpha}. $$
If $\alpha>1$: $1-\alpha<0$, so $t^{1-\alpha}\to0$, giving $\dfrac1{\alpha-1}<\infty$. If $\alpha<1$: $1-\alpha>0$, so $t^{1-\alpha}\to\infty$: the integral diverges. For $\alpha=1$: $\int_1^\infty\frac{dx}x=\lim_{t\to\infty}\ln t=\infty$.
Step 3 · Apply the integral test
The series converges iff the integral converges, i.e. iff $\alpha>1$; it diverges for $0<\alpha\le1$, and trivially diverges for $\alpha\le0$.
Summary Table
| Value of $\alpha$ | Conclusion |
|---|---|
| $\alpha>1$ | Converges, sum $\int_1^\infty x^{-\alpha}dx=\dfrac1{\alpha-1}$ |
| $\alpha=1$ | Diverges (harmonic series) |
| $0<\alpha<1$ | Diverges |
| $\alpha\le0$ | Diverges (terms don't $\to0$) |
Additional Notes
- This is the most fundamental convergence threshold in the whole theory of series — nearly every comparison test in this collection (Problems 82, 84, 99, and many others) ultimately reduces to checking against this exact $p$-series criterion.
- The value $\int_1^\infty x^{-\alpha}dx=\frac1{\alpha-1}$ for $\alpha>1$ gives only a lower bound on the sum $\sum\frac1{n^\alpha}$ (since the integral compares to the tail starting from $n=2$), not the exact value; the true sum is given by the Riemann zeta function $\zeta(\alpha)$, which has no elementary closed form except at even integers.
- Problem 95 immediately extends this same technique by inserting a logarithmic factor $(\ln n)^\alpha$, using the substitution $u=\ln x$ to reduce back to essentially the same integral computation.
Result
The $p$-series converges if and only if $\alpha>1$: