Variable Bases Raised to the Index

Why Variable Bases Need Care

A fixed base raised to a growing power is often straightforward to handle in the root test. A different situation arises when the base itself changes with the index and approaches one. In expressions such as $(1+u/n)^{n^2}$, the base approaches one, but the exponent grows at the same time. The power does not simply approach one: its behavior depends on both changes together.

The root test is especially useful when a term contains such a power. Taking the $n$th root of an exponent proportional to $n^2$ reduces that exponent to one proportional to $n$. The standard exponential limit can then evaluate the resulting expression.

Key idea: When the base is $1+u/n$, a power proportional to $n$ has a finite, generally nontrivial limit. In particular, $(1+u/n)^n$ approaches $e^u$, not $1$.

The Exponential Limit Behind the Method

For a fixed real number $u$, the expression $1+u/n$ is positive for all sufficiently large $n$. The standard exponential limit is

\lim_{n\to\infty}\left(1+\frac{u}{n}\right)^n=e^u.

A useful extension allows the exponent to have a constant multiple of $n$, together with a fixed additive term. For fixed real numbers $\alpha$, $\beta$, and $u$,

\lim_{n\to\infty}\left(1+\frac{u}{n}\right)^{\alpha n+\beta}=e^{\alpha u}.

The fixed term $\beta$ does not affect the limit. The part of the exponent that grows like $n$ determines the limiting value.

Limit rule: For fixed real $u,\alpha,\beta$, with $1+u/n>0$ for all sufficiently large $n$, $$ \left(1+\frac{u}{n}\right)^{\alpha n+\beta}\longrightarrow e^{\alpha u}. $$

Why the Limit Rule Works

We can justify the rule by taking logarithms. For $u\ne 0$, the logarithm of the expression is

\begin{aligned} \ln\left[\left(1+\frac{u}{n}\right)^{\alpha n+\beta}\right] &=(\alpha n+\beta)\ln\left(1+\frac{u}{n}\right)\\ &=\left(\alpha u+\frac{\beta u}{n}\right) \frac{\ln(1+u/n)}{u/n}. \end{aligned}

As $n$ tends to infinity, $\beta u/n$ tends to zero, and the standard logarithmic limit $\ln(1+x)/x\to 1$ as $x\to0$ gives

\lim_{n\to\infty}(\alpha n+\beta)\ln\left(1+\frac{u}{n}\right)=\alpha u.

Since the exponential function is continuous, exponentiating this limit yields $e^{\alpha u}$. If $u=0$, the expression is exactly $1$, and the same formula gives $e^0=1$. This proves the rule in all cases.

Now consider a typical term with a variable-base power, an exponential factor, and possibly a power of $n$:

a_n=\frac{C^n(1+u/n)^{\alpha n^2+\beta n}}{n^p}, \qquad C>0.

For sufficiently large $n$, the variable base is positive, so the $n$th root can be taken directly. The root-test quantity becomes

\sqrt[n]{|a_n|} =C\left(1+\frac{u}{n}\right)^{\alpha n+\beta}n^{-p/n}.

The variable-base factor tends to $e^{\alpha u}$. Also, $n^{-p/n}\to1$, because its logarithm is $-p\ln(n)/n$, which tends to zero. Thus the root-test limit is $Ce^{\alpha u}$. This calculation shows why the exponent's quadratic term matters: taking the $n$th root turns $\alpha n^2$ into $\alpha n$, precisely the scale at which the exponential limit applies.

Using the Root Test

For a series $\sum a_n$, the root test examines the limit $L=\lim_{n\to\infty}\sqrt[n]{|a_n|}$, when that limit exists. If $L<1$, the series converges absolutely. If $L>1$, the terms do not tend to zero, so the series diverges. If $L=1$, the test gives no conclusion.

In applications, simplify the $n$th root before deciding which case applies. In particular, do not discard a variable-base power merely because its base approaches one.

Worked Example: A Series That Converges

Apply the root test to the series whose terms are

a_n=\frac{(1+1/n)^{n^2}}{10^n}.

The terms are positive for every $n\in\mathbb N$, so $\sqrt[n]{|a_n|}=\sqrt[n]{a_n}$. Taking the $n$th root gives

\begin{aligned} \sqrt[n]{a_n} &=\frac{\left((1+1/n)^{n^2}\right)^{1/n}}{(10^n)^{1/n}}\\ &=\frac{(1+1/n)^n}{10}. \end{aligned}

By the exponential limit with $u=1$,

L=\lim_{n\to\infty}\sqrt[n]{a_n} =\frac{e}{10}.

Since $e<10$, we have $L<1$. The root test therefore proves that the series converges absolutely.

Check: Raising the simplified root back to the $n$th power recovers the original term: $\big((1+1/n)^n/10\big)^n=(1+1/n)^{n^2}/10^n=a_n$. The limit factor is also consistent with the logarithmic calculation $n\ln(1+1/n)\to1$, so $(1+1/n)^n\to e$.

Worked Example: A Series That Diverges

Consider the terms

a_n=\frac{(1+2/n)^{n^2}}{5^n}.

Taking the $n$th root gives

\begin{aligned} \sqrt[n]{|a_n|} &=\frac{\left((1+2/n)^{n^2}\right)^{1/n}}{(5^n)^{1/n}}\\ &=\frac{(1+2/n)^n}{5}. \end{aligned}

Here $u=2$, so the variable-base factor tends to $e^2$. Consequently,

L=\lim_{n\to\infty}\sqrt[n]{|a_n|} =\frac{e^2}{5}>1,

since $e^2$ is approximately $7.389$. The root test proves that the series diverges.

Check: The proposed root, when raised to the $n$th power, gives $\big((1+2/n)^n/5\big)^n=(1+2/n)^{n^2}/5^n=|a_n|$. Taking logarithms of its variable-base factor gives $n\ln(1+2/n)\to2$, confirming that this factor tends to $e^2$, not to $1$.

Worked Example: A Limit of One

Now consider terms with both a higher-order exponent and a polynomial factor:

a_n=\frac{(1+1/n)^{2n^2+5n}}{(e^2)^n n^2}.

Taking the $n$th root reduces the exponent $2n^2+5n$ to $2n+5$:

\begin{aligned} \sqrt[n]{|a_n|} &=\frac{(1+1/n)^{2n+5}}{e^2 n^{2/n}}\\ &=\frac{(1+1/n)^{2n}(1+1/n)^5}{e^2 n^{2/n}}. \end{aligned}

The first factor in the numerator tends to $e^2$, while $(1+1/n)^5\to1$. Also, $n^{2/n}\to1$, since its logarithm $2\ln(n)/n$ tends to zero. Therefore,

L=\lim_{n\to\infty}\sqrt[n]{|a_n|} =\frac{e^2\cdot1}{e^2\cdot1}=1.

The root test is inconclusive: its limit is exactly one. This calculation does not prove either convergence or divergence.

Check: Raising the displayed root to the $n$th power gives $(1+1/n)^{2n^2+5n}/(e^{2n}n^2)$, the original positive term. The limiting numerator and exponential denominator both contribute $e^2$ to the root-test limit, while the polynomial factor contributes $1$, confirming the value $L=1$.

A Common Mistake

A frequent error is to let the base tend to one and then treat the entire variable-base power as though it tended to one. For example, in the second worked example, the base $1+2/n$ does tend to one. But the root-test expression contains $(1+2/n)^n$, not just $1+2/n$.

Incorrect calculation: Since $1+2/n\to1$, conclude that $(1+2/n)^n\to1$, and hence that the root-test limit in the second example is $1/5<1$.

This reasoning is wrong because the exponent also changes without bound. The standard exponential limit gives $(1+2/n)^n\to e^2$, so the correct root-test limit is $e^2/5>1$. The correct approach is to evaluate the whole power using the exponential limit, rather than taking the limit of the base alone.

A Reliable Calculation Pattern

When a root-test term contains a variable base raised to an exponent involving $n$, the following sequence keeps the calculation organized:

Take the $n$th root of the full term.
Apply the root to every factor, including exponential and polynomial factors.

Simplify the exponents.
For example, dividing $n^2+5n$ by $n$ leaves $n+5$.

Evaluate each limiting factor.
Use $(1+u/n)^{\alpha n+\beta}\to e^{\alpha u}$, and handle factors such as $n^{p/n}$ separately.

Compare the resulting limit with one.
A limit below one proves absolute convergence; a limit above one proves divergence; a limit of one is inconclusive.

The method matters because two expressions with bases both approaching one can yield different root-test limits when their exponents grow at different rates or have different coefficients. The scale of the exponent must be preserved through the $n$th-root step. Once that is done, the exponential limit turns the coupled base-and-exponent behavior into an ordinary constant that can be compared with one.

Check Your Understanding

For each question, show how the relevant variable-base factor behaves after taking the $n$th root.

What is the limit of $(1+4/n)^{3n-2}$?
If $a_n=(1+2/n)^{3n^2}/8^n$, what is $\lim_{n\to\infty}\sqrt[n]{a_n}$, and what does the root test conclude?
Why does $(1+u/n)^n$ not have limit one merely because $1+u/n\to1$?
What does the root test conclude when its limit is exactly one?
For fixed $p$, why does the factor $n^{-p/n}$ tend to one?

Variable Bases Raised to the Index

What You'll Learn

Why Variable Bases Need Care

The Exponential Limit Behind the Method

Why the Limit Rule Works