Sums With Competing Growth Scales

Why competing scales need an order

A summand can be built from several positive terms that decay at different rates. For example, one contribution might have a smaller power of \(n\) in its denominator but a larger power of \(\log n\). Which contribution determines whether the series converges?

The comparison has a reliable order. First compare the powers of \(n\): a smaller power in the denominator gives slower decay, and eventually outweighs any fixed logarithmic factor. Only when the powers of \(n\) match do the logarithmic exponents decide which term decays more slowly. For a finite sum of positive terms, the slowest-decaying group controls the convergence classification.

Power-logarithmic terms and the dominant pair

Consider a finite sum of terms of the form \(c_i n^{-p_i}(\log n)^{-q_i}\), where \(c_i>0\) and \(p_i,q_i\) are real numbers. We take \(n\ge 2\), so \(\log n>0\). The exponent \(p_i\) describes the power scale; \(q_i\) describes the logarithmic scale.

The order of comparison matters. If \(p_i<p_j\), the \(i\)-term is dominant relative to the \(j\)-term, regardless of the values of \(q_i\) and \(q_j\). If the power exponents are equal, then the smaller logarithmic exponent gives the slower decay. For instance, \(1/(n\log n)\) decays more slowly than \(1/(n(\log n)^2)\).

If several terms share the dominant pair, they all contribute at leading order. Their positive coefficients add; they do not cancel.

Why the dominant terms control the sum

Let the summand be a finite sum \(\displaystyle a_n=\sum_{i=1}^{m}c_i n^{-p_i}(\log n)^{-q_i}\), where \(c_i>0\). Let \((p_*,q_*)\) be its dominant pair, and let \(C_*\) be the sum of the coefficients of all terms with that pair. The leading behavior is

Here, \(a_n\sim d_n\) means \(a_n/d_n\to 1\) as \(n\to\infty\). To justify the claim, divide every term by \(n^{-p_*}(\log n)^{-q_*}\). A term with the same pair contributes its coefficient. Every other term contributes a factor that tends to zero.

Consequently, \(a_n/[C_* n^{-p_*}(\log n)^{-q_*}]\to 1\). In particular, this ratio is bounded above and below by positive constants for all sufficiently large \(n\). The series with terms \(a_n\) therefore has the same convergence behavior as the series with terms \(n^{-p_*}(\log n)^{-q_*}\), by the limit comparison test for positive terms.

Worked examples

A common mistake: comparing logarithms before powers

Consider the positive summand \(1/[n(\log n)^2]+1/(n\log n)\). A tempting but incorrect choice is to call \(1/[n(\log n)^2]\) dominant because it has the larger logarithmic denominator. That reverses the comparison: a larger logarithmic denominator makes a term smaller, not larger.

Therefore \(1/(n\log n)\) is the dominant term. The associated series diverges, so the series of the positive sum diverges too. The correction is to compare equal powers of \(n\) by selecting the smaller logarithmic exponent. More generally, do not use a logarithmic factor to overturn a strict difference in power exponents: powers of \(n\) are compared first.

Why positivity matters

The argument depends on the terms being positive. In a positive finite sum, contributions cannot cancel, and a dominant group remains visible in the total. If signs are allowed, leading contributions can cancel. For example, \(n^{-1}-n^{-1}+n^{-2}\) has a summand equal to \(n^{-2}\), not a summand behaving like \(n^{-1}\). Identifying scales term by term without accounting for cancellation would give the wrong leading behavior.

For the positive sums considered here, the process is systematic: list the scale pairs, choose the smallest power, then the smallest logarithmic exponent among those tied at that power. If multiple terms share that pair, combine their coefficients. Finally, classify the benchmark using its power and logarithmic thresholds. This turns a sum with several apparent growth scales into one standard comparison.