Why Small Arguments Matter
Expressions involving logarithms and exponentials often look more complicated than the terms that determine their long-run size. When \(x\) is close to zero, however, both \(\log(1+x)\) and \(e^x-1\) have the same leading-order behavior as \(x\). This lets us replace a complicated positive term by a simpler comparison term when studying a series.
The key is not to treat the expressions as exactly equal. Instead, we show that their ratio approaches \(1\). That precise relationship is called asymptotic equivalence, and it preserves convergence or divergence for positive-term series.
Meaning of Asymptotic Equivalence
Let \((a_n)\) and \((b_n)\) be sequences with \(b_n\ne 0\) for all sufficiently large \(n\). We say that \(a_n\) is asymptotically equivalent to \(b_n\), and write \(a_n\sim b_n\), if
This means that the ratio of the two terms becomes arbitrarily close to \(1\). It does not mean that \(a_n=b_n\), nor does it mean that their difference tends to zero is enough by itself. For positive-term series, if \(a_n\sim b_n\) and both sequences are positive for all sufficiently large \(n\), then the series \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. This is the series behavior of asymptotically equivalent positive sequences.
In what follows, \(x\) is a real number near zero. The logarithm \(\log(1+x)\) is defined when \(x>-1\), and the quotient defining its equivalence is considered only when \(x\ne0\). The exponential expression \(e^x-1\) is defined for every real \(x\), and its quotient by \(x\) is likewise considered for \(x\ne0\).
Deriving the Logarithmic Equivalent
Apply the Mean Value Theorem to \(f(t)=\log(1+t)\) between \(0\) and \(x\), where \(x>-1\) and \(x\ne0\). The function is continuous on the interval between these points and differentiable in its interior. Its derivative is \(f'(t)=1/(1+t)\). Therefore, for some number \(c\) between \(0\) and \(x\),
Because \(\log(1)=0\), dividing by \(x\) gives
As \(x\to0\), the intermediate value \(c\), which lies between \(0\) and \(x\), also tends to zero. Hence \(1/(1+c)\to1\). This proves the logarithmic equivalent.
Deriving the Exponential Equivalent
Now apply the Mean Value Theorem to \(g(t)=e^t\) between \(0\) and \(x\), with \(x\ne0\). The function is continuous and differentiable, and \(g'(t)=e^t\). Thus for some \(c\) between \(0\) and \(x\),
Since \(e^0=1\), dividing by \(x\) gives \((e^x-1)/x=e^c\). As \(x\to0\), \(c\to0\), so \(e^c\to1\). This proves the exponential equivalent.
These limits apply whether \(x\) approaches zero from above or below. For series, however, we generally choose \(x=x_n>0\), so \(\log(1+x_n)\) and \(e^{x_n}-1\) are positive and can be compared directly with positive power terms.
Substituting a Sequence for the Small Argument
If \(x_n\ne0\) and \(x_n\to0\), the two limits just proved can be applied along the sequence. In particular, if \(x_n>0\), then
Thus \(\log(1+x_n)\sim x_n\) and \(e^{x_n}-1\sim x_n\). The argument sequence \(x_n\) determines the leading order. If \(x_n\) itself behaves like a power of \(1/n\), the logarithmic or exponential term has that same leading order.
Worked Example: A Logarithmic Series with a Convergent Comparison
Determine whether \(\sum_{n=1}^{\infty}\log(1+3/n^2)\) converges. Set \(x_n=3/n^2\). Then \(x_n>0\), \(x_n\to0\), and the logarithmic equivalent applies. Compute the ratio to the comparison term \(1/n^2\):
Here \(3/n^2\to0\), so the second factor tends to \(1\). Therefore the complete ratio tends to \(3\):
The terms are positive, and the limit is a finite positive constant. By limit comparison, the series has the same behavior as \(\sum 1/n^2\), which converges because it is a \(p\)-series with \(p=2>1\). Hence the logarithmic series converges.
Check: The equivalent predicts \(\log(1+3/n^2)\sim3/n^2\). Dividing this predicted leading term by \(1/n^2\) gives \(3\), agreeing with the computed ratio limit.
Worked Example: A Logarithmic Series with a Divergent Comparison
Determine whether \(\sum_{n=1}^{\infty}\log(1+1/n)\) converges. Take \(x_n=1/n\). This sequence is positive and tends to zero. The ratio to the harmonic term is
By the logarithmic limit, this ratio tends to \(1\). Both the logarithmic terms and \(1/n\) are positive. Limit comparison therefore gives the same behavior as the harmonic series \(\sum 1/n\), which diverges. Thus \(\sum_{n=1}^{\infty}\log(1+1/n)\) diverges.
Check: The ratio of the original term to \(1/n\) approaches \(1\), a finite positive constant. The comparison is with the divergent harmonic series, so the conclusion is divergence, not convergence.
Exponential Terms in Positive Series
The exponential equivalent works in the same way. If \(x_n>0\) and \(x_n\to0\), then \(e^{x_n}-1\sim x_n\). A denominator or other positive factor can also be included in the comparison, provided the ratio is computed correctly.
Worked Example: An Exponential Series Compared with a Convergent Power Series
Determine whether \(\sum_{n=1}^{\infty}(e^{1/n}-1)/n\) converges. The terms are positive. Compare them with \(1/n^2\) by forming the ratio:
Since \(1/n\to0\), the exponential limit shows that this ratio tends to \(1\). The given series therefore has the same behavior as \(\sum 1/n^2\), which converges. Hence the exponential series converges.
Check: The ratio calculation leaves exactly \((e^{1/n}-1)/(1/n)\), whose limit is \(1\). The comparison term is \(1/n^2\), not \(1/n\), because the original expression includes the additional factor \(1/n\).
Sign Matters: The Expression \(1-e^{-x}\)
The equivalent \(e^x-1\sim x\) can be used with a negative argument, but the signs must be handled consistently. If \(t>0\) tends to zero, then \(u=-t\) tends to zero from below, and
Multiplying both sides of this leading-order relationship by \(-1\) gives \(1-e^{-t}\sim t\). In particular, \(1-e^{-t}\) is positive. A mistaken sign can turn a positive quantity into a negative comparison and lead to an invalid conclusion.
Worked Example: Checking the Sign in an Exponential Series
Determine whether \(\sum_{n=1}^{\infty}(1-e^{-1/n})\) converges. Every term is positive because \(e^{-1/n}<1\). To compare with \(1/n\), compute the ratio directly:
Set \(u_n=-1/n\). Then \(u_n\to0\) and \(1/n=-u_n\), so this ratio can also be written as
The exponential limit gives \((e^{u_n}-1)/u_n\to1\). Since \((e^{u_n}-1)/(-u_n)\to-1\), the displayed ratio tends to \(-(-1)=1\). Thus \(1-e^{-1/n}\sim1/n\). By limit comparison with the divergent harmonic series, the given series diverges.
Check: The ratio is positive for every \(n\), and its limit is \(1\), consistent with both the exact terms and the predicted equivalent. The series diverges because its positive terms are asymptotic to \(1/n\).
A Common Sign Error
A common incorrect calculation begins with the positive ratio \((1-e^{-1/n})/(1/n)\), then writes
This equality is false: the left side is positive, while the right side is negative. The numerator \(1-e^{-1/n}\) is the negative of \(e^{-1/n}-1\), but the denominator \(1/n\) is positive; it must not also be changed to \(-1/n\) without accounting for that change.
The correct rearrangement is
The inner quotient tends to \(-1\), because its denominator is \(-u_n\) rather than \(u_n\). The leading minus sign then makes the full limit \(1\). A quick sign check before taking a limit often catches this error immediately.
When These Equivalents Are Useful
These equivalents turn expressions with logarithms or exponentials into familiar comparison terms. For example, when \(x_n>0\) and \(x_n\to0\), a term \(\log(1+x_n)\) or \(e^{x_n}-1\) can be compared with \(x_n\). If \(x_n\) is a constant multiple of \(1/n^p\), the corresponding series can often be classified using the \(p\)-series benchmark.
The ratio still needs to be formed carefully. Include every factor in the original term, and compare the resulting ratio with the intended model. As the examples show, a factor of \(1/n\) outside an exponential difference changes a comparison with \(1/n\) into one with \(1/n^2\).
Check Your Understanding
For each question, identify the small argument and state the comparison ratio or leading-order equivalent before deciding the series behavior.
- What is the limit of \(\log(1+x)/x\) as \(x\to0\), and what condition on \(x\) is needed for the logarithm to be defined?
- Determine whether \(\sum_{n=1}^{\infty}\log(1+2/n^3)\) converges.
- Determine whether \(\sum_{n=1}^{\infty}(e^{1/n^2}-1)\) converges.
- What is the correct equivalent of \(1-e^{-x}\) as \(x\to0\) through positive values?
- Explain why a positive-term series whose terms are asymptotically equivalent to \(1/n\) diverges.