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: $$ \sum_{n=1}^{\infty} \left[e^3 - \left(1+\frac{1}{n^2}\right)^{3n^2}\right]^p. $$
Explanation of Method
Since $\left(1+\frac1{n^2}\right)^{3n^2}\to e^3$, the bracketed quantity tends to $0$ — but at what rate matters. Expanding $3n^2\ln(1+\frac1{n^2})$ to second order in $\frac1{n^2}$ shows the base converges to $e^3$ like $e^3\left(1-\frac3{2n^2}\right)$, so the bracket behaves like $\frac{3e^3}{2n^2}$. Raising to the power $p$ gives an overall order of $n^{-2p}$, reducing to a $p$-series comparison with threshold at $p=\frac12$.
- Recall the exponential limit $\left(1+\frac1{n^2}\right)^{n^2}\to e$.
- Expand the logarithm to find the precise rate of convergence to $e^3$.
- Raise to power $p$ and apply the limit comparison test to find the threshold in $p$.
Step 1 · The exponential limit
Since $\left(1+\frac1{n^2}\right)^{n^2}\to e$: $\left(1+\frac1{n^2}\right)^{3n^2} = \left[\left(1+\frac1{n^2}\right)^{n^2}\right]^3 \to e^3$.
Step 2 · Find the precise rate
Using $\ln(1+x)=x-\frac{x^2}2+O(x^3)$ with $x=\frac1{n^2}$: $3n^2\ln\left(1+\frac1{n^2}\right) = 3-\frac3{2n^2}+O\!\left(\frac1{n^4}\right)$, so:
$$ \left(1+\frac1{n^2}\right)^{3n^2} = e^{3-\frac3{2n^2}+O(1/n^4)} = e^3\left(1-\frac3{2n^2}\right)+O\!\left(\frac1{n^4}\right). $$
Therefore:
$$ e^3-\left(1+\frac1{n^2}\right)^{3n^2} \sim \frac{3e^3}{2n^2}. $$
Step 3 · Raise to power $p$ and apply the $p$-series criterion
$$ \left[e^3-\left(1+\frac1{n^2}\right)^{3n^2}\right]^p \sim \left(\frac{3e^3}2\right)^p\cdot\frac1{n^{2p}}. $$
The limit comparison ratio against $b_n=n^{-2p}$ is $\left(\frac{3e^3}2\right)^p$, positive and finite for every $p$. So the series has the same behavior as $\sum\frac1{n^{2p}}$, convergent iff $2p>1$, i.e. $p>\frac12$.
Summary Table
| Quantity | Value |
|---|---|
| Bracket asymptotics | $e^3-(1+1/n^2)^{3n^2} \sim \dfrac{3e^3}{2n^2}$ |
| Effective exponent | $2p$ |
| Convergence | iff $p>\dfrac12$ |
Additional Notes
- The critical threshold $p=\frac12$ arises because the bracket already decays quadratically in $n$ before any power $p$ is applied — raising to power $p$ scales that quadratic exponent to $2p$, so even a modest $p$ just above $\frac12$ is enough to cross the convergence boundary.
- This problem is structurally similar to Problem 40, where an exponential-minus-one expression raised to power $p$ produced a clean $p$-series comparison — here the "base rate" is $n^{-2}$ instead of $n^{-1}$, shifting the critical $p$ from $1$ down to $\frac12$.
- The specific constant $3$ in $e^3$ and $3n^2$ only affects the constant multiplier $\left(\frac{3e^3}2\right)^p$ in front of the asymptotic term, never the convergence threshold itself, which depends purely on the exponent structure ($n^2$ inside, giving order $n^{-2}$).
Result
The convergence of the series depends only on $p$: