Limit of a product (x^k + 1) at infinity
For a fixed positive integer $n$, evaluate the limit: $$ L = \lim_{x \to +\infty} \frac{(x+1)(x^2+1)\cdots(x^n+1)}{\left((nx)^n+1\right)^{\frac{n+1}{2}}}. $$
Explanation of Method
Both the numerator and the denominator are products/powers of expressions that grow like a power of $x$ as $x \to +\infty$, with the "$+1$" terms becoming negligible compared to the dominant power. The strategy is to replace each factor $x^k+1$ by its dominant term $x^k$, add up the resulting exponents in the numerator, and compare that total exponent to the one produced in the denominator. If the two total exponents of $x$ match (as they do here), the limit is a finite constant depending only on $n$.
- Approximate each numerator factor $x^k+1 \approx x^k$ for large $x$, and multiply the exponents.
- Approximate the denominator's base $(nx)^n+1 \approx (nx)^n$, then raise to the power $\frac{n+1}{2}$.
- Compare the exponents of $x$ in numerator and denominator; they agree, so what remains is a constant.
Step 1 · Approximate the numerator
The numerator is the product $(x+1)(x^2+1)\cdots(x^n+1)$. For large $x$, each factor $x^k+1$ is dominated by $x^k$, since the constant $1$ becomes negligible:
$$ (x+1)(x^2+1)\cdots(x^n+1) \approx x^1\cdot x^2 \cdots x^n = x^{1+2+\cdots+n} = x^{\frac{n(n+1)}{2}}, $$
using the formula $1+2+\cdots+n = \dfrac{n(n+1)}{2}$ for the sum of the first $n$ positive integers.
Step 2 · Approximate the denominator
The denominator is $\left((nx)^n+1\right)^{\frac{n+1}{2}}$. For large $x$, $(nx)^n$ dominates the added $1$:
$$ (nx)^n + 1 \approx (nx)^n = n^n x^n. $$
Raising to the power $\frac{n+1}{2}$:
$$ \left((nx)^n+1\right)^{\frac{n+1}{2}} \approx \left(n^n x^n\right)^{\frac{n+1}{2}} = n^{\frac{n(n+1)}{2}} \, x^{\frac{n(n+1)}{2}}. $$
Step 3 · Compare exponents and take the limit
Both the numerator and denominator have grown to the same power of $x$, namely $x^{\frac{n(n+1)}{2}}$, so this factor cancels in the ratio:
$$ L \approx \lim_{x \to +\infty} \frac{x^{\frac{n(n+1)}{2}}}{n^{\frac{n(n+1)}{2}}\, x^{\frac{n(n+1)}{2}}} = \frac{1}{n^{\frac{n(n+1)}{2}}}. $$
Since the neglected "$+1$" terms become relatively smaller and smaller compared to the dominant powers as $x \to +\infty$, this approximation becomes exact in the limit.
Summary Table
| Piece | Dominant behavior as $x \to +\infty$ |
|---|---|
| Numerator | $x^{\frac{n(n+1)}{2}}$ |
| Denominator | $n^{\frac{n(n+1)}{2}} x^{\frac{n(n+1)}{2}}$ |
| Ratio | $\dfrac{1}{n^{\frac{n(n+1)}{2}}}$ |
Additional Notes
- The matching total exponent $\frac{n(n+1)}{2}$ in both numerator and denominator is exactly why the limit is a finite, nonzero constant rather than $0$ or $\infty$ — this is the same idea as comparing the degrees of two polynomials in a rational function.
- The result depends only on $n$: for $n=1$ it gives $\frac{1}{1^{1}}=1$; for $n=2$ it gives $\frac{1}{2^{3}}=\frac{1}{8}$; and so on, growing rapidly smaller as $n$ increases.
- This kind of "leading term" comparison is a standard shortcut for limits at infinity: whenever every factor differs from a pure power of $x$ by a lower-order additive term, only the leading powers determine the limit.
Result
The value of the limit is: