Investigate the convergence of product from n=1
Investigate the convergence of $\displaystyle\prod_{n=1}^{\infty}\left(1+\frac1n\right)^n n^{1/n}$.
Explanation of Method
A necessary condition for any infinite product $\prod a_n$ to converge is $a_n\to1$. Here $a_n=(1+1/n)^n\cdot n^{1/n}\to e\cdot1=e\ne1$, so the product cannot converge — it must diverge to $+\infty$, since every factor is positive and eventually close to $e>1$.
- Compute the limit of the general term $a_n$.
- Invoke the necessary condition $a_n\to1$ for convergence.
- Confirm via the logarithmic series that $P_N$ grows like $e^N$.
Step 1 · The limit of the general term
$$ \left(1+\frac1n\right)^n\longrightarrow e, \qquad n^{1/n}\longrightarrow1 \qquad(n\to\infty), $$ so $a_n=\left(1+\frac1n\right)^nn^{1/n}\to e\cdot1=e$.
Step 2 · The necessary condition fails
An infinite product $\prod a_n$ (with all $a_n>0$) can only converge to a nonzero limit if $a_n\to1$. Since $a_n\to e\ne1$, the product necessarily diverges.
Step 3 · Rate of divergence
$$ \ln a_n = n\ln\left(1+\frac1n\right)+\frac{\ln n}n = \left(1-\frac1{2n}+O\!\left(\frac1{n^2}\right)\right)+\frac{\ln n}n \longrightarrow1. $$ So $\ln a_n\to1$, meaning $\sum_{n=1}^N\ln a_n\sim N\to\infty$, hence $$ \boxed{P_N=\exp\left(\sum_{n=1}^N\ln a_n\right)\sim e^N\longrightarrow+\infty} $$ (Numerical check: $P_1=2$, $P_2=9\sqrt2/2\approx6.36$, $P_3\approx21.7$ — growing roughly by a factor of $e\approx2.718$ at each step, as predicted.)
Summary Table
| Property | Value |
|---|---|
| $\lim_{n\to\infty}a_n$ | $e$ |
| Growth rate | $P_N\sim e^N$ |
| Conclusion | diverges to $+\infty$ |
Additional Notes
- This is the cleanest possible illustration of the necessary condition $a_n\to1$: no delicate series comparison is needed at all, since the terms visibly approach $e\ne1$ rather than merely failing a borderline convergence test.
- Contrast with Problem 371, where $a_n=1+n^{-\alpha}\to1$ for every $\alpha$, and the finer question of series-convergence-rate is what decides convergence; here that finer analysis is unnecessary because the coarse necessary condition already fails.
- The factor $n^{1/n}\to1$ contributes nothing to the divergence — it is entirely driven by $(1+1/n)^n\to e$, so this product diverges at essentially the same rate as $\prod(1+1/n)^n$ alone would.