Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=1}^{\infty} (\sqrt[n]{a}-1), \qquad a>1. $$
Explanation of Method
Although $a^{1/n}\to1$, the rate at which it approaches $1$ matters. Expanding $a^{1/n}=e^{(\ln a)/n}$ via the exponential's derivative at $0$ shows $a^{1/n}-1\sim\frac{\ln a}{n}$ — exactly the harmonic rate. Since $\ln a>0$ for $a>1$, the limit comparison test against $\sum1/n$ shows the series diverges.
- Show $a^{1/n}\to1$ using logarithms.
- Find the precise rate via the exponential-derivative limit $\lim_{h\to0}\frac{a^h-1}{h}=\ln a$.
- Apply the limit comparison test against $\sum1/n$.
Step 1 · Show $a^{1/n}\to1$
$\ln(a^{1/n})=\frac{\ln a}{n}\to0$, so $a^{1/n}=e^{(\ln a)/n}\to e^0=1$; hence $a^{1/n}-1\to0$, a necessary but not sufficient condition for convergence.
Step 2 · Find the precise asymptotic rate
Setting $h=\frac1n$:
$$ \lim_{n\to\infty} n(a^{1/n}-1) = \lim_{h\to0^+}\frac{a^h-1}{h} = \left.\frac{d}{dx}a^x\right|_{x=0} = \ln a. $$
Since $a>1$, $\ln a>0$, so this limit is positive and finite, confirming $a^{1/n}-1\sim\dfrac{\ln a}{n}$.
Step 3 · Apply the limit comparison test
With $b_n=\dfrac1n$:
$$ \lim_{n\to\infty}\frac{a^{1/n}-1}{1/n} = \ln a > 0. $$
Since $\sum\frac1n$ diverges (harmonic series), by the limit comparison test $\sum_{n=1}^{\infty}(a^{1/n}-1)$ diverges as well.
Summary Table
| Quantity | Value |
|---|---|
| $a^{1/n}-1$ asymptotics | $\sim\dfrac{\ln a}{n}$ |
| Limit comparison ratio (vs. $1/n$) | $\ln a>0$ |
| Conclusion | Diverges |
Additional Notes
- The limit $\lim_{h\to0}\frac{a^h-1}h=\ln a$ is exactly the definition of the derivative of $a^x$ at $x=0$, so this problem is really an application of differentiability of the exponential function in disguise.
- Since $\ln a>0$ strictly for every $a>1$, the divergence conclusion holds uniformly for the entire range $a>1$ — the specific value of $a$ only rescales the terms, never changes the qualitative outcome.
- Compare this to Problem 17 ($\sum 1/\sqrt[n]{n+\alpha}$), where the $n$-th root similarly tends to $1$ but the terms don't vanish at all (limit $=1\ne0$); here the terms do vanish, but too slowly (like $1/n$) to converge.
Result
The series diverges: