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: $$ \sum_{n=1}^{\infty} \left(e^{\frac{1}{n+1}} - 1\right)^p. $$
Explanation of Method
Since $e^x-1\sim x$ for small $x$, the base $e^{1/(n+1)}-1$ behaves like $\frac1{n+1}$, and raising to the power $p$ gives an overall order of $(n+1)^{-p}\sim n^{-p}$. This reduces the problem to an ordinary $p$-series comparison, settled entirely by whether $p>1$.
- Expand $e^{1/(n+1)}-1$ via its Taylor series.
- Raise to the power $p$ and apply the limit comparison test against $\sum1/(n+1)^p$.
- State the resulting $p$-series criterion.
Step 1 · Taylor expand
$$ e^{\frac1{n+1}}-1 = \frac1{n+1}+\frac1{2(n+1)^2}+O\!\left(\frac1{(n+1)^3}\right) \sim \frac1{n+1}. $$
Step 2 · Raise to power $p$ and apply the limit comparison test
With $b_n=\dfrac1{(n+1)^p}$: setting $h=\frac1{n+1}\to0^+$ and using $\lim_{h\to0}\frac{e^h-1}h=1$,
$$ \lim_{n\to\infty} \frac{\left(e^{1/(n+1)}-1\right)^p}{1/(n+1)^p} = \left[\lim_{n\to\infty}(n+1)\left(e^{1/(n+1)}-1\right)\right]^p = 1^p = 1. $$
Positive and finite, so $\sum\left(e^{1/(n+1)}-1\right)^p$ shares the convergence behavior of $\sum\dfrac1{(n+1)^p}$.
Step 3 · Apply the $p$-series criterion
$\sum\frac1{(n+1)^p}$ is a shifted $p$-series, convergent iff $p>1$. Hence:
$$ \sum_{n=1}^{\infty} \left(e^{1/(n+1)}-1\right)^p \quad\text{converges iff } p>1,\ \text{diverges iff } p\le1. $$
Summary Table
| Quantity | Value |
|---|---|
| $e^{1/(n+1)}-1$ asymptotics | $\sim\dfrac1{n+1}$ |
| Limit comparison ratio (vs. $1/(n+1)^p$) | $1$ |
| Convergence | iff $p>1$ |
Additional Notes
- The identity $\lim_{h\to0}\frac{e^h-1}h=1$ used here is the exponential analogue of the sine limit used in Problems 38–39; both are instances of a function's derivative at $0$ controlling the leading-order asymptotic behavior of a related series.
- This problem is the exponential counterpart to Problem 34's root-based series: there, $a^{1/n}-1\sim\frac{\ln a}{n}$ gave a fixed effective exponent of $1$ regardless of $a$; here the exponent $p$ is a free parameter, so the threshold $p=1$ becomes the tunable convergence boundary instead.
- Since $(n+1)^{-p}$ and $n^{-p}$ are asymptotically identical multiples of each other, shifting the index by $1$ (writing $n+1$ instead of $n$) has no effect on the convergence criterion, only on the precise value of any resulting sum.
Result
The convergence of the series depends only on $p$: