Fixed Shifts and Constant Multipliers

Why Fixed Changes Often Leave the Order Unchanged

In a series term, changing $n$ to $n+4$, or replacing $n$ by $7n$, may look like a substantial alteration. For power and logarithmic comparison models, however, these changes do not alter the term's asymptotic order. Establishing the ratio precisely makes that fact usable in series comparisons.

There is an important qualification: a fixed shift does not preserve the order of every possible sequence. The results here concern standard power and logarithmic models, and constant multiples of their terms. We will state the model and the relevant conditions explicitly.

Definition: For eventually positive sequences $a_n$ and $b_n$, write $a_n\sim b_n$ if $\lim_{n\to\infty}a_n/b_n=1$. Say they have the same asymptotic order if $\lim_{n\to\infty}a_n/b_n=L$ for some $0<L<\infty$. Asymptotic equivalence implies the same order, but the converse need not hold.

For instance, $3b_n$ has the same asymptotic order as $b_n$, since their ratio is $3$, but $3b_n\not\sim b_n$. The distinction matters when stating a ratio limit: a change in the index by a fixed additive amount often gives equivalence, while scaling the index generally introduces a constant.

The Power-Logarithm Model

For real numbers $p$ and $q$, consider the positive model

M_n=n^{-p}(\log n)^{-q},\qquad n>1.

The exponents may be positive, zero, or negative. The requirement $n>1$ ensures that $\log n>0$, so the real power of $\log n$ is defined and positive. Fix a real shift $h$ and a positive scaling factor $\lambda$. For all sufficiently large $n$, both $n+h>1$ and $\lambda n>1$, so the corresponding model terms are defined.

Theorem: For fixed $h\in\mathbb R$, fixed $\lambda>0$, and fixed real $p,q$, $$ \frac{M_{n+h}}{M_n}\longrightarrow 1, \qquad \frac{M_{\lambda n}}{M_n}\longrightarrow \lambda^{-p}. $$ Consequently, a fixed shift gives an asymptotically equivalent term, while positive scaling of the index gives the same asymptotic order, with ratio constant $\lambda^{-p}$.

The expression $M_{\lambda n}$ is evaluated from the same model formula, even when $\lambda n$ is not an integer. In applications, this notation describes expressions such as $(\lambda n)^{-p}(\log(\lambda n))^{-q}$.

Why the Ratio Limits Hold

The power part and the logarithmic part can be compared separately. This is legitimate because the model is a product of positive factors. For a shift, the power ratio is $(1+h/n)^{-p}$, which tends to $1$. The logarithmic ratio also tends to $1$: the logarithms of $n+h$ and $n$ have ratio tending to $1$. Raising a positive ratio tending to $1$ to a fixed real power still gives a limit of $1$.

\begin{aligned} \frac{M_{n+h}}{M_n} &= \left(\frac{n+h}{n}\right)^{-p} \left(\frac{\log(n+h)}{\log n}\right)^{-q}\\ &= \left(1+\frac{h}{n}\right)^{-p} \left(\frac{\log(n+h)}{\log n}\right)^{-q}. \end{aligned}

Indeed, $(n+h)/n=1+h/n\to1$. For the logarithmic factor, write

\log(n+h)=\log n+\log\left(1+\frac{h}{n}\right).

Since $\log(1+h/n)\to0$ and $\log n\to\infty$, it follows that $\log(n+h)/\log n\to1$. Both factors in the ratio therefore tend to $1$, proving the shift limit.

For scaling, the power ratio is exactly $\lambda^{-p}$, while the logarithmic ratio tends to $1$:

\begin{aligned} \frac{M_{\lambda n}}{M_n} &= \left(\frac{\lambda n}{n}\right)^{-p} \left(\frac{\log(\lambda n)}{\log n}\right)^{-q}\\ &= \lambda^{-p} \left(1+\frac{\log\lambda}{\log n}\right)^{-q} \longrightarrow \lambda^{-p}. \end{aligned}

Here $\log(\lambda n)=\log n+\log\lambda$, and $\log\lambda/\log n\to0$. This proves both claims, including the constant that remains when the index is scaled.

Worked Examples

Worked Example: A Fixed Shift in a Power

Compare $a_n=1/(n+4)^3$ with $b_n=1/n^3$. Both are positive for $n\in\mathbb N$. Compute their ratio:

\frac{a_n}{b_n} = \frac{1/(n+4)^3}{1/n^3} = \frac{n^3}{(n+4)^3} = \left(\frac{n}{n+4}\right)^3 = \left(\frac{1}{1+4/n}\right)^3.

As $n\to\infty$, $4/n\to0$, so $a_n/b_n\to1$. Thus $a_n\sim b_n$. The series $\sum_{n=1}^{\infty} b_n=\sum_{n=1}^{\infty}1/n^3$ converges, because it is a $p$-series with $p=3>1$. The limit comparison test applies to these eventually positive terms and gives convergence of $\sum a_n$ as well.

Check the ratio calculation in the limit: the numerator and denominator in $n/(n+4)$, after division by $n$, give $1/(1+4/n)\to1$. The ratio limit is therefore $1$, as claimed.

Worked Example: Scaling the Index in a Power

Let $a_n=1/(4n+3)^2$ and compare it with $b_n=1/n^2$. Factor $n$ out of $4n+3$:

\begin{aligned} \frac{a_n}{b_n} &= \frac{1/(4n+3)^2}{1/n^2} = \frac{n^2}{(4n+3)^2}\\ &= \left(\frac{n}{4n+3}\right)^2 = \left(\frac{1}{4+3/n}\right)^2 \longrightarrow \frac{1}{16}. \end{aligned}

The limit is positive and finite, so the sequences have the same asymptotic order, though they are not asymptotically equivalent. The comparison series $\sum 1/n^2$ converges because $p=2>1$. By limit comparison, $\sum 1/(4n+3)^2$ converges.

For an explicit check of the constant, $3/n\to0$, so $4+3/n\to4$, and its reciprocal squared tends to $1/4^2=1/16$. Dropping the factor $4$ would incorrectly give a ratio limit of $1$; the convergence conclusion would happen to be unchanged, but the asymptotic comparison would be misstated.

Worked Example: A Shift in a Logarithmic Model

For $n\ge2$, let $$ a_n=\frac{1}{(n+5)(\log(n+5))^2}, \qquad b_n=\frac{1}{n(\log n)^2}. $$

Form the ratio and separate its power and logarithmic parts:

\begin{aligned} \frac{a_n}{b_n} &= \frac{n(\log n)^2}{(n+5)(\log(n+5))^2}\\ &= \frac{n}{n+5} \left(\frac{\log n}{\log(n+5)}\right)^2. \end{aligned}

The first factor tends to $1$, since $n/(n+5)=1/(1+5/n)$. Also, $\log(n+5)/\log n\to1$, by the fixed-shift argument, so its reciprocal squared tends to $1$. Therefore $a_n/b_n\to1$.

The series $\sum_{n=2}^{\infty}1/(n(\log n)^2)$ converges by the logarithmic threshold test for this model. The terms are positive, and their ratio tends to $1$, so limit comparison proves that $\sum_{n=2}^{\infty}a_n$ converges. As a check, the calculated ratio is a product of two factors each tending to $1$; hence its limit is indeed $1$.

Using Ratio Limits in Series Comparisons

The limit comparison test applies when $a_n$ and $b_n$ are positive for all sufficiently large $n$, and

\lim_{n\to\infty}\frac{a_n}{b_n}=L,\qquad 0<L<\infty.

Under these conditions, $\sum a_n$ and $\sum b_n$ either both converge or both diverge. A finite number of initial terms does not affect convergence, so eventual positivity is enough. When a fixed shift gives ratio limit $1$, or a positive scaling gives a finite positive constant such as $\lambda^{-p}$, this test transfers the known convergence behavior of the comparison series.

A constant multiplying the terms themselves behaves similarly. If $A>0$, then

\frac{A a_n}{b_n} = A\frac{a_n}{b_n}.

Thus, if $a_n/b_n\to L$ with $0<L<\infty$, then $A a_n/b_n\to AL$, also positive and finite. Multiplying every term of a positive series by a fixed positive constant therefore preserves convergence or divergence. This is different from scaling the index: for $M_n=n^{-p}(\log n)^{-q}$, replacing $n$ by $\lambda n$ produces the factor $\lambda^{-p}$ in the ratio.

A Pitfall: Shifts Do Not Preserve Every Sequence

The shift theorem proved above is a claim about the power-logarithm model, not a universal rule for arbitrary sequences. For a concrete counterexample, define $u_n=1$ when $n$ is even and $u_n=2$ when $n$ is odd. The ratio $u_{n+1}/u_n$ alternates:

\frac{u_{n+1}}{u_n} = \begin{cases} 2,& n\text{ even},\\[2pt] 1/2,& n\text{ odd}. \end{cases}

This ratio has no limit, so $u_{n+1}$ is not asymptotically equivalent to $u_n$, and the two sequences do not have the same order under the finite-positive-ratio definition. The incorrect calculation would be to assert “a fixed shift has ratio limit $1$” for every sequence. The correction is to identify the sequence model and prove its ratio limit; the result applies to $n^{-p}(\log n)^{-q}$, but the alternating example shows why it cannot be assumed for an arbitrary sequence.

There is a second condition to retain in series applications: the comparison terms must be eventually positive. A ratio limit of zero or infinity is not the finite positive limit required by the limit comparison test. Nor can a zero coefficient be treated as a harmless positive multiplier. State the positivity and ratio-limit conditions before drawing a convergence conclusion.

Practical Checklist

Identify the model.
Check whether the terms are built from powers of $n$, logarithms, or both; do not assume the shift rule for an arbitrary sequence.

Write the ratio.
Put the candidate term over a positive comparison term, then separate powers and logarithms where useful.

Keep constant factors.
A fixed additive shift gives ratio limit $1$ for the stated model; scaling $n$ by $\lambda$ gives $\lambda^{-p}$.

Check the test's hypotheses.
For limit comparison, verify eventual positivity and a finite, strictly positive ratio limit before transferring the series behavior.

The practical benefit is that an awkward shift or fixed coefficient often does not require a new convergence calculation. Once its ratio to a standard model has a positive finite limit, that model determines the series behavior through limit comparison. The ratio still needs to be computed correctly: equal order does not always mean asymptotic equivalence.

Check Your Understanding

Use a ratio calculation in each question, and state any conditions needed for a series comparison.

For $p=4$ and fixed $h=7$, what is the limit of $(n+h)^{-p}/n^{-p}$?
For $M_n=n^{-2}(\log n)^{-1}$, what is the limit of $M_{3n}/M_n$?
If $a_n/b_n\to5$ and both sequences are eventually positive, what does limit comparison say about their series?
Why does $1/(4n+3)^2$ have the same asymptotic order as $1/n^2$ but is not asymptotically equivalent to it?
Give a reason that the fixed-shift result for power-logarithm models cannot be applied automatically to every sequence.