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: $$ \sum_{n=1}^{\infty} \left(a^{1/n} + a^{-1/n} - 2\right), \qquad a>0. $$
Explanation of Method
Unlike Problem 34, here both $a^{1/n}$ and $a^{-1/n}$ appear together. Expanding each via the exponential Taylor series, the $\pm\ln a/n$ linear terms exactly cancel when added, leaving a leading term of order $1/n^2$ instead of $1/n$ — a much faster decay that lands the series in convergent territory.
- Handle the trivial case $a=1$ separately.
- Expand $a^{1/n}+a^{-1/n}$ via Taylor series, noting the linear-term cancellation.
- Apply the limit comparison test against $\sum1/n^2$.
Step 1 · The case $a=1$
If $a=1$, every term is $1+1-2=0$, so the series trivially converges to $0$. Assume $a>0,\;a\ne1$ for the rest.
Step 2 · Taylor expand and cancel linear terms
Writing $a^{\pm1/n}=e^{\pm(\ln a)/n}$ and expanding to second order:
$$ a^{1/n}+a^{-1/n} = \left(1+\frac{\ln a}{n}+\frac{(\ln a)^2}{2n^2}\right) + \left(1-\frac{\ln a}{n}+\frac{(\ln a)^2}{2n^2}\right) + O\!\left(\frac1{n^4}\right) = 2+\frac{(\ln a)^2}{n^2}+O\!\left(\frac1{n^4}\right). $$
The odd (linear) terms cancel exactly, so:
$$ a^{1/n}+a^{-1/n}-2 \sim \frac{(\ln a)^2}{n^2}. $$
Step 3 · Apply the limit comparison test
With $b_n=\dfrac1{n^2}$:
$$ \lim_{n\to\infty} \frac{a^{1/n}+a^{-1/n}-2}{1/n^2} = (\ln a)^2 > 0 \quad (\text{since } a\ne1). $$
Since $\sum\frac1{n^2}$ converges ($p=2>1$), by the limit comparison test $\sum_{n=1}^{\infty}(a^{1/n}+a^{-1/n}-2)$ converges too.
Summary Table
| Case | Result |
|---|---|
| $a=1$ | Every term is $0$; converges trivially |
| $a>0,\ a\ne1$ | $a^{1/n}+a^{-1/n}-2\sim(\ln a)^2/n^2$; converges |
Additional Notes
- The cancellation of the linear ($1/n$) terms is the key mechanism here, contrasting sharply with Problem 34: adding a quantity and its reciprocal-exponent counterpart symmetrizes away the leading-order divergent contribution, upgrading the decay rate from $1/n$ to $1/n^2$.
- The quantity $a^{1/n}+a^{-1/n}-2$ can also be written as $\left(a^{1/(2n)}-a^{-1/(2n)}\right)^2$, making its non-negativity for all $a>0$ immediately visible — every term of the series is $\ge0$.
- This convergent behavior holds for every $a>0$ (including $a$ between $0$ and $1$, where $\ln a<0$ but $(\ln a)^2$ is still positive), so the result is symmetric under $a\mapsto1/a$.
Result
The series converges for every $a>0$: