Shifted p-Series and Constant Multipliers

Introduction

A series can look different from a familiar p-series because its denominator contains a fixed shift or its terms have a constant multiplier. For example, $6/(n+3)^2$ is not written in the standard form $1/n^p$. To classify it, the key question is whether those changes alter the behavior of the tail: the terms remaining after any finite number of initial terms.

A fixed shift changes the denominator by a bounded amount as $n$ grows, and a positive constant multiplier rescales the terms. Neither changes the p-series threshold. The argument rests on precise inequalities, not merely on the impression that two expressions look similar.

Definition and Convergence Threshold

Definition: Let $C>0$, let $a$ be a fixed real number, and let $n_0$ be an integer such that $n+a>0$ for every integer $n\ge n_0$. A shifted p-series with constant multiplier is $$ \sum_{n=n_0}^{\infty}\frac{C}{(n+a)^p}, $$ where $p$ is real. The condition $n+a>0$ ensures that the base of the power is positive.

The standard p-series threshold is that $\sum 1/n^p$ converges when $p>1$ and diverges when $p\le 1$. We will use this established result and show that a fixed shift and a positive constant leave that threshold unchanged.

Convergence concerns the sequence of partial sums. A series converges if its partial sums approach a finite limit; for a series with positive terms, its partial sums increase, so divergence means they are unbounded. The terms with $n$ below any fixed cutoff form only a finite sum. Adding or removing finitely many finite terms cannot change whether the remaining partial sums are bounded.

Why a Fixed Shift Does Not Change the Tail

For sufficiently large $n$, the shifted quantity $n+a$ lies between two positive constant multiples of $n$. In particular, whenever $n\ge 2|a|$, we have $$ \frac{n}{2}\le n+a\le \frac{3n}{2}. $$ For $a=0$ these inequalities hold as well. If $p>0$, taking reciprocal $p$-th powers reverses the order of the positive quantities, giving $$ \left(\frac{2}{3}\right)^p\frac{1}{n^p} \le \frac{1}{(n+a)^p} \le 2^p\frac{1}{n^p}. $$ Multiplying by $C>0$ preserves the inequalities. Thus, for all sufficiently large $n$, $$ C\left(\frac{2}{3}\right)^p\frac{1}{n^p} \le \frac{C}{(n+a)^p} \le C\,2^p\frac{1}{n^p}. $$ Both constants are positive and fixed. The shifted terms are therefore bounded above and below by constant multiples of the corresponding p-series terms.

These inequalities explain both directions of the threshold. If $p>1$, the upper bound is a constant multiple of the convergent p-series $\sum 1/n^p$, so the shifted series has bounded partial sums and converges. If $0<p\le 1$, the lower bound is a positive constant multiple of a divergent p-series, so the shifted partial sums are unbounded and the series diverges.

When $p=0$, each term is $C$, so the partial sums increase without bound. When $p<0$, write $-p=q$ with $q>0$. For all sufficiently large $n$, $n+a\ge1$, and then $$ \frac{C}{(n+a)^p}=C(n+a)^q\ge C. $$ The partial sums again increase without bound.

Theorem: For $C>0$, fixed real $a$, and valid indices $n\ge n_0$ with $n+a>0$, the series $$ \sum_{n=n_0}^{\infty}\frac{C}{(n+a)^p} $$ converges if and only if $p>1$. It diverges for $p\le1$.

Worked Examples

Worked Example: A Shifted Series with Exponent Greater Than One

Classify $$ \sum_{n=1}^{\infty}\frac{6}{(n+3)^2}. $$ Here $C=6$, $a=3$, and $p=2$. The first five terms are finite: $$ \frac{6}{4^2}=\frac{3}{8},\qquad \frac{6}{5^2}=\frac{6}{25},\qquad \frac{6}{6^2}=\frac{1}{6},\qquad \frac{6}{7^2}=\frac{6}{49},\qquad \frac{6}{8^2}=\frac{3}{32}. $$ For $n\ge6$, $n+3\le 3n/2$, since this inequality is equivalent to $3\le n/2$. Hence $$ \frac{6}{(n+3)^2} \le \frac{6}{(3n/2)^2} =\frac{6}{9n^2/4} =\frac{8}{3n^2} \le \frac{24}{n^2}. $$ The series $\sum 1/n^2$ converges because its exponent is $2>1$. Therefore the tail partial sums of the given series are bounded above by a constant multiple of convergent p-series partial sums. Including the five finite initial terms preserves convergence. The series converges.

Check: The bound uses $n+3\le3n/2$, so its reciprocal square is bounded above, not below, by $4/(9n^2)$. This gives the displayed upper bound and supports the convergence conclusion.

Worked Example: The Harmonic Threshold with a Shift

Classify $$ \sum_{n=1}^{\infty}\frac{5}{n+2}. $$ Here $C=5$, $a=2$, and $p=1$. For $n\ge4$, the inequality $n+2\le3n/2$ holds because it is equivalent to $2\le n/2$. Since the denominators are positive, taking reciprocals gives $$ \frac{1}{n+2}\ge\frac{2}{3n}. $$ Multiplication by $5$ gives $$ \frac{5}{n+2}\ge\frac{10}{3n}. $$ The partial sums from $n=4$ onward are consequently at least $10/3$ times the corresponding partial sums of the harmonic series. Those harmonic partial sums are unbounded, so the shifted series diverges. The first three terms are finite and do not affect the conclusion.

Check: Both inequalities preserve their direction as written: $n+2\le3n/2$ gives $1/(n+2)\ge2/(3n)$, and multiplying by the positive number $5$ gives the stated lower bound.

Worked Example: A Fractional Exponent at or Below the Threshold

Classify $$ \sum_{n=1}^{\infty}\frac{2}{\sqrt{n+1}}. $$ This has $C=2$, $a=1$, and $p=1/2$. For $n\ge2$, $n+1\le3n/2$, so $$ \frac{1}{\sqrt{n+1}} \ge \frac{1}{\sqrt{3n/2}} =\sqrt{\frac{2}{3}}\frac{1}{\sqrt n}. $$ Multiplying by $2$ gives $$ \frac{2}{\sqrt{n+1}} \ge 2\sqrt{\frac{2}{3}}\frac{1}{\sqrt n}. $$ The constant $2\sqrt{2/3}$ is positive, and the p-series $\sum 1/\sqrt n$ diverges because $p=1/2\le1$. The shifted series therefore diverges.

Check: Squaring the positive quantities in $n+1\le3n/2$ is unnecessary; taking square roots preserves the inequality, and reciprocating then reverses it. The resulting lower bound has the correct direction for proving divergence.

Worked Example: A Nonpositive Exponent

Classify $$ \sum_{n=1}^{\infty}\frac{4}{(n+2)^{-1}}. $$ Here $p=-1$, so the term simplifies to $$ \frac{4}{(n+2)^{-1}}=4(n+2). $$ For every $n\ge1$, $4(n+2)\ge4$. Thus the first $N$ terms have sum at least $4N$, which grows without bound as $N$ increases. The series diverges.

Check: The negative exponent moves the power to the numerator: $1/(n+2)^{-1}=n+2$, not $1/(n+2)$. The terms grow rather than shrink.

A Common Comparison Mistake

When comparing a term with a familiar p-series, keep the constant multiplier in the inequality. For positive $n$, it is true that $1/(n+4)<1/n$. But multiplying this inequality by $3$ gives $$ \frac{3}{n+4}<\frac{3}{n}, $$ not $3/(n+4)<1/n$. The latter claim is false: at $n=4$, $3/(n+4)=3/8$, while $1/n=1/4$, and $3/8>1/4$.

That mistaken upper bound would not establish convergence anyway. To prove divergence of $\sum 3/(n+4)$, use a lower bound instead. For $n\ge8$, $n+4\le3n/2$, and therefore $$ \frac{3}{n+4}\ge\frac{3}{3n/2}=\frac{2}{n}. $$ The harmonic partial sums are unbounded, so the partial sums of this shifted series are unbounded as well. The multiplier $3$ changes the scale of the terms, but it does not turn a divergent harmonic tail into a convergent one.

How to Classify These Series Reliably

For a series of the form $C/(n+a)^p$, the exponent determines the threshold, while the shift and positive multiplier affect only the size of the terms. A careful classification can be organized as follows:

Check the parameters.
Confirm that $C>0$, the shift $a$ is fixed, and $n+a>0$ on the specified indices.

Identify the exponent.
Compare $p$ with $1$. The threshold is $p>1$ for convergence; $p\le1$ gives divergence.

Justify the tail behavior.
For $p>0$, bound $n+a$ between fixed positive multiples of $n$ once $n$ is large enough. For $p\le0$, show directly that the terms are bounded below by a positive constant.

Account for the initial terms.
Any terms before the tail estimate begins are finite in number and finite in value, so they do not alter convergence or divergence.

This reasoning matters because it separates the feature that controls the classification—the exponent—from changes that only affect the tail by fixed factors. It also prevents an invalid comparison: a term may be smaller than $1/n^p$ while still being larger than $C/n^p$ for a multiplier $C>1$. When proving convergence, establish an appropriate upper bound; when proving divergence, establish an appropriate positive lower bound.

Check Your Understanding

For each question, state the relevant exponent and justify the conclusion using the p-series threshold or an explicit tail bound.

Does $\sum_{n=1}^{\infty} 7/(n+5)^3$ converge or diverge?
Why can removing the first four terms of a shifted p-series not change its convergence classification?
For $p>0$, what inequalities hold between $1/(n+a)^p$ and constant multiples of $1/n^p$ when $n\ge2|a|$?
Classify $\sum_{n=1}^{\infty} 2/(n+1)^{2/3}$, and identify a useful lower bound for sufficiently large $n$.
What is the error in claiming $4/(n+3)<1/n$ merely because $1/(n+3)<1/n$?