Limit of the product of (1 + k/n²)
Evaluate the limit, where $n$ is a positive integer: $$ L = \lim_{n \to \infty} \prod_{k=1}^n\left(1+\frac{k}{n^2}\right). $$
Explanation of Method
Each individual factor $1+\frac{k}{n^2}$ tends to $1$ as $n\to\infty$ (since $\frac{k}{n^2}\le\frac1n\to0$ for every $k$ from $1$ to $n$), but there are $n$ such factors multiplied together, so the product is not automatically $1$ — this is a "many factors, each close to $1$" situation, best handled by taking logarithms to turn the product into a sum. Approximating each logarithm by $\ln(1+u)\approx u$ turns the sum into $\sum_{k=1}^n\frac k{n^2}$, which is evaluated exactly using the formula for the sum of the first $n$ positive integers.
- Take the logarithm of the product, converting it into a sum.
- Approximate each term of the sum using $\ln(1+u)\approx u$.
- Evaluate the resulting sum using $\sum_{k=1}^nk=\frac{n(n+1)}2$.
- Take the limit of the sum and exponentiate to recover $L$.
Step 1 · Take the logarithm of the product
Let $P_n = \prod_{k=1}^n\left(1+\frac k{n^2}\right)$. Taking logarithms converts the product into a sum:
$$ \ln P_n = \sum_{k=1}^n \ln\left(1+\frac{k}{n^2}\right). $$
Step 2 · Approximate each logarithm
For every $k$ from $1$ to $n$, $\frac k{n^2}\le\frac n{n^2}=\frac1n\to0$ as $n\to\infty$, so every term is uniformly small. Using $\ln(1+u)\approx u$ for small $u$:
$$ \ln P_n \approx \sum_{k=1}^n \frac{k}{n^2}. $$
Step 3 · Evaluate the sum exactly
Using the formula for the sum of the first $n$ positive integers, $\sum_{k=1}^nk=\dfrac{n(n+1)}2$:
$$ \sum_{k=1}^n\frac{k}{n^2} = \frac1{n^2}\cdot\frac{n(n+1)}2 = \frac{n+1}{2n}. $$
Step 4 · Take the limit and exponentiate
As $n\to\infty$:
$$ \lim_{n\to\infty}\ln P_n = \lim_{n\to\infty}\frac{n+1}{2n} = \lim_{n\to\infty}\left(\frac12+\frac1{2n}\right) = \frac12. $$
Since $\exp$ is continuous:
$$ L = \lim_{n\to\infty} P_n = \lim_{n\to\infty} e^{\ln P_n} = e^{\lim_{n\to\infty}\ln P_n} = e^{1/2} = \sqrt e. $$
Summary Table
| Quantity | Expression | Limit as $n\to\infty$ |
|---|---|---|
| $\ln P_n$ | $\approx\displaystyle\sum_{k=1}^n\frac{k}{n^2}=\frac{n+1}{2n}$ | $\dfrac12$ |
| $P_n$ | $e^{\ln P_n}$ | $\sqrt e$ |
Additional Notes
- The approximation $\ln(1+u)\approx u$ introduces an error of order $u^2$ for each term; summing these errors over all $n$ terms gives a total error of order $\sum_{k=1}^n\left(\frac{k}{n^2}\right)^2 = O\!\left(\frac1n\right)$ (using $\sum k^2=O(n^3)$), which still vanishes as $n\to\infty$ — confirming the approximation is rigorous, not just heuristic, in the limit.
- This is a standard technique for limits of large products where every factor tends to $1$: take logarithms to turn the product into a (Riemann-sum-like) sum, and evaluate that sum either in closed form (as here, via $\sum k=\frac{n(n+1)}2$) or via an integral approximation for more complicated sums.
- The final answer $\sqrt e$ reflects that the product's "total logarithmic growth," $\sum\frac k{n^2}$, settles at exactly $\frac12$ — geometrically, this sum is a Riemann-sum approximation to $\int_0^1t\,dt=\frac12$ under the substitution $t=\frac kn$, offering yet another way to see why the answer is $e^{1/2}$.
Result
The value of the limit is: