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Series and Partial Sums · Tutorial 567 of 998

Shifted Reciprocal Differences

You will learn to write partial sums for shifted reciprocal differences, locate the terms that do not cancel, and use the vanishing tail to find the series value.

Beginner 9 min read

What You'll Learn

  • Set up partial sums for differences of reciprocals with integer shifts.
  • Identify the initial block and the moving tail left after cancellation.
  • Derive a general formula for the partial sums when one shift is larger.
  • Bound the tail to justify the infinite-series value.
  • Check finite sums and avoid common index-shift errors.

Introduction

A reciprocal difference can look as though its terms should cancel, but the cancellation is not always immediate. For example, the positive term \(1/n\) and the negative term \(-1/(n+2)\) have matching reciprocals two places apart. Writing several terms reveals the pattern, but the partial sum must still be tracked carefully: some terms remain at the beginning, while others remain at the far end.

This tutorial develops a reliable way to evaluate such series. The central task is to write a finite partial sum first, identify exactly which terms cancel, and only then take the limit of the partial sums.

Key idea: When the two reciprocals have fixed integer shifts, cancellation leaves a fixed block at the beginning and a block that moves to larger denominators. The moving block tends to zero, so the fixed block determines the series value.

Definition and Indexing

Consider a series whose \(n\)th term is the difference of two shifted reciprocals:

$$ \sum_{n=1}^{\infty}\left(\frac{1}{n+r}-\frac{1}{n+s}\right), $$

where \(r\) and \(s\) are fixed nonnegative integers. The \(N\)th partial sum is the sum of the first \(N\) terms:

$$ S_N=\sum_{n=1}^{N}\left(\frac{1}{n+r}-\frac{1}{n+s}\right). $$

The infinite series converges to a value \(S\) if the sequence of partial sums has a finite limit, in which case \(S=\lim_{N\to\infty}S_N\). In this topic, the shift difference is fixed: the offsets do not change as \(n\) changes.

Suppose first that \(r<s\). The positive reciprocals have denominators \(r+1,r+2,\ldots,r+N\). The subtracted reciprocals have denominators \(s+1,s+2,\ldots,s+N\). The two lists overlap, but the second list starts later. After cancellation, the positive terms from \(r+1\) through \(s\) remain at the beginning, and the negative terms from \(N+r+1\) through \(N+s\) remain at the end.

The Partial-Sum Formula

Theorem: If \(r\) and \(s\) are nonnegative integers with \(r<s\), then for every positive integer \(N\), $$ S_N=\sum_{k=r+1}^{s}\frac{1}{k}-\sum_{k=N+r+1}^{N+s}\frac{1}{k}. $$ Consequently, the series converges and $$ \sum_{n=1}^{\infty}\left(\frac{1}{n+r}-\frac{1}{n+s}\right) =\sum_{k=r+1}^{s}\frac{1}{k}. $$

The formula separates the partial sum into two parts: a fixed initial block and a finite tail whose denominators increase with \(N\). The proof uses only cancellation of matching terms and a bound on that tail.

1
Write both finite sums.
The positive part is \(\sum_{k=r+1}^{N+r}1/k\), and the negative part is \(\sum_{k=s+1}^{N+s}1/k\). These index ranges follow by setting \(k=n+r\) and \(k=n+s\), respectively.
2
Separate the cancelled terms.
Because \(r<s\), the overlapping indices run from \(s+1\) through \(N+r\) when \(N+r\ge s+1\). Cancelling this common block leaves the initial positive block \(r+1,\ldots,s\) and the final negative block \(N+r+1,\ldots,N+s\). The resulting formula also holds when the ranges do not overlap, since the same finite-sum identity applies.
3
Show that the moving tail vanishes.
There are \(s-r\) terms in the final block, each at most \(1/(N+r+1)\). Thus its sum is between zero and \((s-r)/(N+r+1)\), which tends to zero as \(N\) tends to infinity.

Therefore the partial sums approach the fixed initial block, proving the stated series value. If \(r=s\), every term is zero and the sum is zero. If \(r>s\), reverse the order of the two shifts: the series is the negative of the case with the smaller shift first. These cases make the role of the order of subtraction explicit.

Worked Examples

Worked Example: A Difference with a Two-Place Shift

Evaluate the first three terms and the infinite series \(\sum_{n=1}^{\infty}(1/n-1/(n+2))\).

First, write the first three terms without skipping indices:

$$ S_3=\left(1-\frac{1}{3}\right) +\left(\frac{1}{2}-\frac{1}{4}\right) +\left(\frac{1}{3}-\frac{1}{5}\right). $$

The individual differences are \(2/3\), \(1/4\), and \(2/15\). With common denominator \(60\), their sum is

$$ S_3=\frac{2}{3}+\frac{1}{4}+\frac{2}{15} =\frac{40}{60}+\frac{15}{60}+\frac{8}{60} =\frac{63}{60} =\frac{21}{20}. $$

Now use the cancellation pattern for a general \(N\):

$$ S_N= \left(1+\frac12+\frac13+\cdots+\frac1N\right) -\left(\frac13+\frac14+\cdots+\frac1{N+2}\right) =1+\frac12-\frac1{N+1}-\frac1{N+2}. $$

For \(N=3\), this formula gives \(3/2-1/4-1/5\). In sixtieths, that is \(90/60-15/60-12/60=63/60=21/20\), agreeing with the direct calculation. As \(N\) grows, the last two terms tend to zero. Therefore the infinite series sums to \(1+1/2=3/2\). The fixed uncancelled terms are \(1\) and \(1/2\), not the terms that appear at the far end of a finite partial sum.

Worked Example: Three Reciprocals Remain

Evaluate \(\sum_{n=1}^{\infty}(1/(n+2)-1/(n+5))\), and check its second partial sum.

The first two terms are

$$ S_2=\left(\frac13-\frac16\right)+\left(\frac14-\frac17\right) =\frac16+\frac{3}{28} =\frac{14}{84}+\frac{9}{84} =\frac{23}{84}. $$

For a general partial sum, the positive denominators run from \(3\) to \(N+2\), while the negative denominators run from \(6\) to \(N+5\). Cancelling the overlap gives

$$ S_N=\frac13+\frac14+\frac15 -\left(\frac1{N+3}+\frac1{N+4}+\frac1{N+5}\right). $$

For \(N=2\), the formula gives \(1/3+1/4+1/5-(1/5+1/6+1/7)\). The \(1/5\) terms cancel, leaving \(1/3+1/4-1/6-1/7\). In eighty-fourths, this is \(28/84+21/84-14/84-12/84=23/84\), confirming the direct sum.

Each of the three tail terms tends to zero as \(N\) increases. The infinite sum is the fixed block:

$$ \frac13+\frac14+\frac15 =\frac{20}{60}+\frac{15}{60}+\frac{12}{60} =\frac{47}{60}. $$

Worked Example: Reversing the Order Reverses the Sign

Evaluate \(\sum_{n=1}^{\infty}(1/(n+4)-1/(n+1))\), and verify its third partial sum.

Here the larger shift occurs in the positive reciprocal. The terms are the negatives of those in the same series with the shifts reversed. The first three terms give

$$ S_3=\left(\frac15-\frac12\right) +\left(\frac16-\frac13\right) +\left(\frac17-\frac14\right) =-\frac3{10}-\frac16-\frac3{28}. $$

Using denominator \(420\), the three fractions are \(-126/420\), \(-70/420\), and \(-45/420\). Hence \(S_3=-241/420\).

For a general \(N\), cancellation leaves a negative initial block and a positive tail:

$$ S_N=-\left(\frac12+\frac13+\frac14\right) +\left(\frac1{N+2}+\frac1{N+3}+\frac1{N+4}\right). $$

At \(N=3\), the tail is \(1/5+1/6+1/7\). Combining this with \(-1/2-1/3-1/4\) gives, in denominator \(420\), \(-210-140-105+84+70+60=-241\), so the result is \(-241/420\), as calculated directly. The tail tends to zero, and the series value is

$$ -\left(\frac12+\frac13+\frac14\right) =-\left(\frac6{12}+\frac4{12}+\frac3{12}\right) =-\frac{13}{12}. $$

A Common Indexing Mistake

A common error is to identify the first uncancelled term correctly but then stop tracking the shift at the far end. For the series \(\sum_{n=1}^{\infty}(1/n-1/(n+2))\), it is incorrect to write \(S_N=1+1/2\) for every finite \(N\). For example, that would claim \(S_3=3/2\), even though direct calculation gives \(21/20\).

The missing terms are the negative tail terms \(1/(N+1)\) and \(1/(N+2)\). The correct finite formula is \(S_N=1+1/2-1/(N+1)-1/(N+2)\). Only after taking the limit can the tail be discarded. A finite partial sum and the limit of the partial sums are related, but they are not generally equal for a finite \(N\).

Check before taking a limit: Write the first few terms and the last few terms of the partial sum. Confirm that the starting indices match the original shifts and that the moving tail has exactly as many terms as the difference between the shifts.

Why the Method Is Useful

The partial-sum approach distinguishes a genuine infinite-series value from a visual impression of cancellation. It shows which terms survive at each finite stage and gives a direct reason that the remaining tail disappears. The size of the shift difference tells you how many terms appear in the fixed block and in the moving tail.

For \(r<s\), the result can be read as follows:

$$ \sum_{n=1}^{\infty}\left(\frac{1}{n+r}-\frac{1}{n+s}\right) =\frac{1}{r+1}+\frac{1}{r+2}+\cdots+\frac{1}{s}. $$

This formula applies because the shifts are fixed nonnegative integers and the series begins at \(n=1\). Changing the starting index changes which reciprocal terms appear in the fixed block, so the index convention must be included in any calculation.

Check Your Understanding

For each question, begin by identifying the finite partial sum and its uncancelled terms.

  1. For \(\sum_{n=1}^{\infty}(1/(n+1)-1/(n+4))\), what fixed terms remain, and what is the series value?
  2. Write \(S_N\) for \(\sum_{n=1}^{\infty}(1/(n+3)-1/(n+5))\), including the moving tail.
  3. For \(\sum_{n=1}^{\infty}(1/(n+5)-1/(n+2))\), determine the sign and value of the series.
  4. Why is it not generally correct to replace a finite partial sum by only its fixed initial block?
  5. How many terms occur in the fixed block when \(r<s\), and how can you show the moving tail tends to zero?

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