Why Indexing Errors Matter
A finite series can have a familiar-looking formula and still be evaluated incorrectly if its indices are misread. Starting an index at \(2\) rather than \(1\) can change the first term; using the upper index as the number of terms can add extra terms; and cancelling a telescoping sum too far can remove a term that should remain.
The safest check is to write out the terms at the beginning and end of the sum. These terms reveal what is actually being added and which terms can cancel. A partial sum provides a second check: it must agree with both the expanded terms and the formula used to evaluate them.
Read the Limits Before Using a Formula
In the finite sum \(\sum_{k=m}^{n} a_k\), the index \(k\) takes each integer value from \(m\) through \(n\), including both endpoints. Thus the sum is \(a_m+a_{m+1}+\cdots+a_n\), and its number of terms is \(n-m+1\), provided \(m\) and \(n\) are integers with \(m\le n\).
The \(+1\) matters because both endpoints are included. For example, the integer indices from \(4\) through \(8\) are \(4,5,6,7,8\): there are \(8-4+1=5\) indices, not four.
For a geometric sum, the first term is the value of the summand at the lower index. If the summand is \(c r^{k-p}\), then the first term in a sum beginning at \(k=m\) is \(c r^{m-p}\), not automatically \(c\). For \(N\) terms with first term \(a\) and common ratio \(r\ne1\), the finite geometric sum is \(a(1-r^N)/(1-r)\). When \(r=1\), the sum is \(Na\).
Why the Term Count Is \(n-m+1\)
To see why the count includes \(+1\), list the indices beginning at \(m\). The first index is \(m\), the second is \(m+1\), and the last is \(n\). If there are \(N\) indices, the last one is also \(m+(N-1)\), since the first index uses up one of the \(N\) positions. Equating this with \(n\) gives \(n=m+N-1\), so \(N=n-m+1\).
This count is also the exponent needed when the ratio is used to build a geometric sum. A list of five terms has powers \(r^0,r^1,r^2,r^3,r^4\): the final exponent is one less than the number of terms. Confusing the upper index with the number of terms can therefore give both the wrong count and the wrong last power.
Worked Example: Find the First Term and Count the Terms
Worked Example: Find the First Term and Count the Terms
Evaluate \(\sum_{k=2}^{6}3\cdot2^{k-1}\), and check the result by expansion.
The index runs from \(2\) through \(6\), so the number of terms is \(6-2+1=5\). The first term is found by substituting \(k=2\), not by taking the coefficient \(3\) alone:
Substituting each index gives the complete sum:
The terms form a geometric sequence with first term \(a=6\), ratio \(r=2\), and \(N=5\) terms. The formula gives:
The formula agrees with the expansion. A check of the endpoints gives \(k=2\) as the first term \(6\), and \(k=6\) as the last term \(3\cdot2^5=96\), confirming that the listed five terms match the index limits.
A Complete Check for Geometric Sums
For a proposed geometric sum, check three things separately: the value at the lower index, the number of included indices, and the value at the upper index. The first and last terms should match the expansion; the count should match the number of terms in that expansion. Once these checks agree, use the geometric sum formula with that first term and count.
Evaluate the summand at the starting index to identify the actual first term.
For inclusive integer limits \(m\) and \(n\), use \(n-m+1\).
Evaluate the summand at \(n\) and confirm that it is the last term in the expansion.
Use the first term, ratio, and term count, then compare the result with the expanded sum when practical.
Worked Example: Correct a Term-Count Error
Worked Example: Correct a Term-Count Error
Evaluate \(\sum_{j=4}^{8}5(1/3)^{j-2}\). A common incorrect count is \(8-4=4\); check the count and evaluate the sum.
The indices are \(4,5,6,7,8\), so there are \(8-4+1=5\) terms. Substituting \(j=4\) gives the first term, and substituting \(j=8\) gives the last:
Writing every term confirms that there are five:
As a formula check, use \(a=5/9\), \(r=1/3\), and \(N=5\):
If we used four terms instead, the result would be \(\frac59+\frac5{27}+\frac5{81}+\frac5{243}=\frac{600}{729}\), omitting the required last term \(5/729\). The expansion, endpoint substitution, and formula all confirm the correct answer.
Track the Endpoints in a Telescoping Sum
A telescoping sum is one in which terms cancel after the summands are expanded. The cancellation should be performed on the expanded terms, not inferred from a vague impression that “most terms disappear.” For instance, in \(\sum_{k=m}^{n}(b_k-b_{k+1})\), expansion gives \(b_m-b_{m+1}+b_{m+1}-b_{m+2}+\cdots+b_n-b_{n+1}\). Each intermediate term appears once positively and once negatively. The first positive term and the final negative term have no matching terms to cancel.
This gives the finite identity:
The upper endpoint of the original index is \(n\), but the final negative term is \(b_{n+1}\). That extra one comes from evaluating \(b_{k+1}\) at \(k=n\). Losing it is a common cancellation-endpoint error.
Why Telescoping Leaves These Terms
The identity follows by writing out the finite sum and grouping its positive and negative terms. For every index \(j\) with \(m+1\le j\le n\), the positive \(b_j\) from the term with \(k=j\) cancels the negative \(-b_j\) from the preceding term with \(k=j-1\). The term \(b_m\) has no preceding negative term in the sum, and \(-b_{n+1}\) has no following positive term.
Equivalently, pair and cancel the terms in the expansion:
The conclusion depends on the exact form \(b_k-b_{k+1}\) and on inclusive limits \(m\) through \(n\). Writing out the endpoint terms is a direct way to catch an off-by-one error.
Worked Example: Check the Cancellation Endpoint
Worked Example: Check the Cancellation Endpoint
Evaluate \(\sum_{k=1}^{5}\left(\frac1k-\frac1{k+1}\right)\), and identify the error in the claim that the result is \(1-\frac15\).
Substitute \(k=1,2,3,4,5\) into the summand. The final negative term comes from \(k=5\), so it is \(-1/6\):
The terms \(1/2,1/3,1/4,\) and \(1/5\) each cancel: each appears once with a positive sign and once with a negative sign. The uncancelled terms are \(1\) and \(-1/6\), so the result is \(5/6\).
The incorrect answer \(1-1/5\) treats \(1/5\) as the final negative term. But the last summand is obtained by substituting \(k=5\), giving \(1/5-1/6\). The value \(1/5\) is an intermediate term, and its positive and negative occurrences cancel. Directly checking the expansion confirms the corrected endpoint and result.
A Concrete Indexing Pitfall
Consider the incorrect calculation for the first worked example: “The coefficient is \(3\), the upper index is \(6\), so use six terms beginning with \(3\).” This fails in two separate ways. At \(k=2\), the summand is \(3\cdot2^1=6\), not \(3\); and the indices \(2\) through \(6\) supply five terms, not six. A calculation can therefore look like a valid geometric sum while representing a different list of terms.
Correcting only one of these mistakes is not enough. The first term and the count must both match the index range. In a telescoping sum, likewise, correcting the first term does not fix an endpoint that has been cancelled incorrectly. Check each feature independently.
Check Your Understanding
For each question, use the index limits to check the terms before evaluating or simplifying.
- How many terms are in \(\sum_{k=3}^{9}a_k\)?
- What is the first term of \(\sum_{j=2}^{6}4\cdot3^{j-1}\), and how many terms are included?
- In \(\sum_{k=2}^{5}(b_k-b_{k+1})\), what are the two terms that remain after cancellation?
- For \(\sum_{k=1}^{4}(1/k-1/(k+1))\), what is the final negative term before cancellation?
- A sum runs from index \(5\) through index \(10\). Is its term count \(10-5\) or \(10-5+1\)? Explain why.
