Telescoping sum repair: keep both boundary pairs

Write the finite sum before cancelling

Telescoping sum repair starts by identifying which terms actually have matching partners. A shift of two positions leaves a different boundary from a shift of one. This original worksheet examines the finite sum of 1 divided by k times (k+2), with k running from 1 through a positive integer N. The calculation uses exact rational arithmetic. It does not assign a value to an infinite series or infer a pattern from measurements.

Call the sum S_N. Its first term is 1/(1 times 3), or 1/3. Its second term is 1/(2 times 4), or 1/8. Its third is 1/(3 times 5), or 1/15. The upper index N tells us how many of these stipulated terms to include. It is not the value of the final denominator, which depends on both k and its shifted partner k+2.

For N=3, direct addition gives 1/3+1/8+1/15. A common denominator is 120: these fractions become 40/120, 15/120, and 8/120. Their sum is 63/120=21/40. This small exact result will serve as an independent check of the telescoping calculation. Saving it before simplifying the general expression makes a missing boundary term easier to detect later.

Verify the term identity and its factor

The useful identity is 1/[k(k+2)]=(1/2)(1/k-1/(k+2)). Combine the two fractions on the right before trusting it. Their difference has numerator (k+2)-k=2 and denominator k(k+2). Multiplying by 1/2 gives numerator 1, which reproduces the assigned term. Because k is a positive integer, both denominators are nonzero. The identity is valid at every index used in the finite sum.

The factor 1/2 belongs to every transformed term. Omitting it would double the assigned sum. For example, at k=1, the difference 1-1/3 equals 2/3, while the assigned first term is 1/3. The factor converts one to the other. A cancellation diagram should retain that factor outside the entire finite bracket, rather than attach it to just the first or last fraction.

Now write S_N=(1/2) times the sum, for k=1 through N, of (1/k-1/(k+2)). This is a finite collection of N identities added together. We can separate its positive and negative terms using ordinary finite distributivity and addition. No assertion about a limit is required. Keeping the finite upper index visible prevents an unproved infinite operation from entering a calculation that needs none.

Align the two index ranges

The positive reciprocals have denominator indices 1, 2, 3, and so on through N. The negative reciprocals have indices 3, 4, 5, and so on through N+2. A reciprocal cancels only if the same denominator index occurs in both lists with opposite signs. The lists start and finish two positions apart. Their first and last displayed entries alone do not describe the whole uncancelled boundary.

For N at least 3, the common indices are 3 through N. Every reciprocal with one of those indices occurs once positively and once negatively. After cancelling those pairs, the positive reciprocals 1 and 1/2 remain. At the far end, the negative reciprocals -1/(N+1) and -1/(N+2) remain. No positive copy of either final index exists in the original positive range.

Thus the finite result is S_N=(1/2)(1+1/2-1/(N+1)-1/(N+2)). The two positive terms come from the beginning of the positive list. The two negative terms come from the end of the shifted negative list. Simplifying 1+1/2 to 3/2 gives the equivalent expression 3/4-1/[2(N+1)]-1/[2(N+2)]. Preserve the index explanation beside either compact formula.

Check the three-term case both ways

At N=3, expand the transformed expression as (1/2) times [(1-1/3)+(1/2-1/4)+(1/3-1/5)]. The positive 1/3 in the last pair cancels the negative 1/3 in the first. The remaining bracket is 1+1/2-1/4-1/5. The reciprocal 1/2 has no negative matching copy here, and 1/4 has no positive matching copy.

Use denominator 20 for the remaining bracket. It becomes 20/20+10/20-5/20-4/20=21/20. Multiplying the entire result by 1/2 gives 21/40, agreeing with direct addition. The cancellation and the direct three-fraction calculation took different routes to the same exact answer. Agreement checks the factor, signs, denominator indices, and surviving terms in this particular example.

A common shortcut keeps just the first positive reciprocal and the final negative reciprocal. That would give (1/2)(1-1/(N+2)). At N=3, it gives (1/2)(1-1/5)=2/5, or 16/40. The direct result is 21/40, so the shortcut is short by 5/40=1/8. The disagreement is not a rounding issue: both values are exact fractions.

Compare the correct and incorrect brackets directly. The shortcut dropped 1/2-1/(N+1). Its missing contribution to the sum is therefore (1/2)(1/2-1/(N+1)). For N=3, that is (1/2)(1/2-1/4)=1/8, exactly the observed shortfall. This locates the lost pair rather than merely replacing an incorrect total with the expected one.

Check short ranges without inventing interior terms

At N=2, the positive index list is 1,2 and the negative list is 3,4. They have no common indices, so there is no interior cancellation to perform. The four-term boundary expression still applies: (1/2)(1+1/2-1/3-1/4). Its value is 11/24. Direct addition gives 1/3+1/8=11/24 as well. An empty interior range does not make the sum empty.

At N=1, there is just one original term, 1/3. Substituting into the compact formula gives (1/2)(1+1/2-1/2-1/3). The displayed positive and negative halves cancel each other, leaving (1/2)(1-1/3)=1/3. The formula remains correct, but its four displayed positions do not represent four distinct surviving indices in this short case. The shared index 2 appears in that representation and cancels.

This short-range check matters when a general explanation says that two terms survive at each end. For N at least 2, those beginning and ending boundary indices are distinct. At N=1, the compact representation overlaps. Treat the overlap explicitly, or state the longer-range condition before describing four distinct boundary terms. A correct formula can coexist with an incomplete explanation of why it works at the smallest allowed input.

For another direct check, N=4 adds the term 1/(4 times 6)=1/24 to the three-term sum. We obtain 21/40+1/24=63/120+5/120=68/120=17/30. The boundary formula gives 3/4-1/10-1/12, also 17/30. This comparison checks the updated right boundary at indices 5 and 6 rather than reusing the N=3 ending fractions.

Keep changed denominators in a separate problem

The adjacent-shift identity 1/[k(k+1)]=1/k-1/(k+1) belongs to a different summand. Its positive index range is 1 through N, and its negative range is 2 through N+1. Only one boundary reciprocal survives at each end, giving 1-1/(N+1). That calculation is correct for its own input. It does not repair the original k(k+2) calculation while preserving the assigned denominators.

At N=3, the adjacent-shift terms are 1/2, 1/6, and 1/12. They add to 3/4. The original terms are 1/3, 1/8, and 1/15, which add to 21/40. Comparing these lists shows two changes at once: the shift changed from two to one, and the partial-fraction identity no longer needs a factor 1/2. Importing only the familiar final boundary pattern would conceal both changes.

A useful repair record keeps three lines together: the assigned term, the verified term identity, and the positive and negative index ranges. Then write the uncancelled terms before reducing them to a compact expression. If an answer fails a small direct check, compare those three lines first. Repeating the same unsupported cancellation cannot identify a lost term when the index lists themselves have never been written down.

Use exact controls within their stated scope

An exact rational control checked N from 1 through 100. For every input, direct addition of 1/[k(k+2)] matched the four-term boundary formula. It also checked that the difference from the one-boundary shortcut equaled (1/2)(1/2-1/(N+1)). These are finite checks over declared inputs. The term identity and matching-index argument explain the result for arbitrary positive N, with the N=1 overlap addressed separately.

There is no need to round either route to decimals to compare them. For N=3, 21/40 and 2/5 already show an exact mismatch. Converting to a short decimal display could make a later comparison depend on a precision choice that the worksheet never needed. Retain the rational result and its unreduced boundary expression together so a reader can inspect the arithmetic and the cancellation structure independently.

The article makes no claim about an infinite sum, convergence, an average of partial sums, or rearranging infinitely many terms. Every displayed sum has a specified finite number of terms. Those other questions require their own definitions and arguments. Here the repair is complete once the finite identity, retained boundaries, short-range cases, and direct checks agree for the assigned summand.

Verify the formula by consecutive differences

There is another finite verification that does not depend on reading a cancellation diagram. Let F_N denote the proposed expression 3/4-1/[2(N+1)]-1/[2(N+2)]. The first value is F_1=3/4-1/4-1/6=1/3, matching the first assigned term. If consecutive values of F_N differ by the next assigned summand, this base case will establish the formula for every positive integer N by induction. Both parts of that check are needed.

For N at least 2, the preceding value is F_(N-1)=3/4-1/(2N)-1/[2(N+1)]. Subtract it from F_N while keeping parentheses around the entire preceding expression. The constants 3/4 cancel, as do the opposite copies of 1/[2(N+1)]. What remains is 1/(2N)-1/[2(N+2)]. Combining these fractions gives [(N+2)-N]/[2N(N+2)], which is exactly 1/[N(N+2)]. That is the assigned term at index N.

The direct sum has the same update: S_N-S_(N-1)=1/[N(N+2)], because every earlier term occurs once in each finite prefix. Starting from S_1=F_1, the matching update gives S_2=F_2, then S_3=F_3, and so on. Induction expresses this reasoning for arbitrary positive N. This check supplies a base value and a precise update rule; matching a few displayed totals alone would supply neither a proof of the update nor an argument covering later indices.

Apply the same test to the incorrect one-boundary expression G_N=(1/2)(1-1/(N+2)). Its consecutive difference is 1/[2(N+1)]-1/[2(N+2)], or 1/[2(N+1)(N+2)]. This differs from the assigned next term 1/[N(N+2)]. The shortcut therefore fails the finite update test as well as the N=3 direct check. A proposed repair should preserve the actual term added when N increases, rather than only resemble the shape of a familiar telescoping answer. Keep the base value beside the update calculation: an expression with the correct consecutive difference could still differ from the intended sum by a fixed constant.

Bound the optional native accounts

The account for Beistle green tinsel garland 31559213 shows a green strand and an inset detail. The photograph does not authenticate the catalog pack count, ply, length, width, material, maker identity, or performance.

The account for Beistle red-white-blue tinsel garland 31559273 shows a strand with an inset detail. Neither detail view proves another supplied strand. Catalog identifiers remain labels; dimensions, delivered quantity, material, maker authenticity, and performance are unverified. Neither account supplies the finite sum or cancellation rule.

Keep neighboring calculations distinct

The alternating series sum mismatch guide distinguishes an ordinary partial-sum limit from limits of averages. This worksheet uses only finite exact addition and cancellation between two explicit index ranges. No alternative averaging operation or infinite value is needed to recover its missing boundary terms.

The linear congruence solutions guide retains both original-modulus classes after reducing an integer relation. Here the retained objects are uncancelled reciprocal terms from shifted finite lists. The modular lifting proof does not establish a telescoping identity; the term algebra and index matching do that work.

Separate any transaction from the calculation

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