I’m cramming for a test on sequences/series and I keep tripping over sigma notation, especially when the index doesn’t start at 1. My brain does this little detour like I’m counting theater seats but someone labeled the first row as Row 0. Anyway, here’s where I’m stuck.
Simple example first: For the arithmetic sum ∑(k=1 to 4) of (2k+1), I did it two ways. Expanding: 3 + 5 + 7 + 9 = 24. Using the formula with first=3, last=9, n=4: S = n/2*(first+last) = 4/2*(3+9) = 24. That lines up, yay.
But then if I switch to ∑(k=0 to 4) of (2k+1), I got confused. I originally set n=4 again, took first=1 and last=9, and did S = 4/2*(1+9) = 20. But expanding gives 1 + 3 + 5 + 7 + 9 = 25. So I guess I messed up the “n” – it should be the number of terms, which from 0 to 4 is 5. I keep forgetting that part.
Related headache: reindexing a geometric sum. For ∑(k=2 to n) of 3*(1/2)^k, I tried: let j = k – 2, so it becomes ∑(j=0 to n-2) of 3*(1/2)^(j+2) = (3/4) * ∑(j=0 to n-2) (1/2)^j. Then I used the geometric formula and ended up with something like (3/2) * [1 – (1/2)^(n-1)]. But I’m not confident about that exponent – is it n-1 or did I miscount the terms again?
My questions:
– How do I systematically figure out the correct “n” (number of terms) in finite sums so I don’t keep making off-by-one mistakes?
– When I reindex (like k → j = k – 2), what’s a reliable way to track the new bounds and the final exponent so it lines up with the geometric sum formula?
– If you can, could you point out exactly where my geometric attempt goes fuzzy?
I’m trying to build a little mental checklist before the test so I stop second-guessing myself. Thanks!
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3 Responses
I feel you on the “Row 0” brain glitch-been there! The safest habit is to count terms first: if k runs from a to b (inclusive), then n = b − a + 1. That’s why Σ(k=0 to 4)(2k+1) has n = 4 − 0 + 1 = 5 terms, giving 1 + 3 + 5 + 7 + 9 = 25; your earlier 20 happened because n was taken as 4 instead of 5. For reindexing, I always do three tiny moves: (1) set your new index, say j = k − c, so k = j + c; (2) transform the bounds: when k = a, j = a − c; when k = b, j = b − c; (3) rewrite the summand in j and factor any constants. In your geometric example Σ(k=2 to n) 3(1/2)^k, let j = k − 2 so j runs 0 to n − 2 and 3(1/2)^k = 3(1/2)^{j+2} = (3/4)(1/2)^j. Now use the standard geometric formula Σ(j=0 to m) r^j = (1 − r^{m+1})/(1 − r) with r = 1/2 and m = n − 2, so Σ(j=0 to n−2) (1/2)^j = 1 − (1/2)^{n−1} all over 1/2, which is 2(1 − (1/2)^{n−1}). Multiplying by 3/4 gives (3/2)(1 − (1/2)^{n−1}). Your exponent n − 1 is exactly right-notice it equals the number of terms after reindexing (from 0 to n − 2 is n − 1 terms), which is a great mental check: for a geometric sum starting at 0, the power on r in the “1 − r^power” is the number of terms. Quick worked check: if n = 3, the sum should be 3(1/2)^2 + 3(1/2)^3 = 3/4 + 3/8 = 9/8; the formula gives (3/2)(1 − (1/2)^2) = (3/2)(3/4) = 9/8. So your arithmetic fix (use n = upper − lower + 1) and your geometric reindexing are both spot on-just anchor yourself with “count terms first,” then “shift, shift bounds, substitute,” and you’ll dodge the off-by-one gremlins.
I feel this in my bones. Any time the index doesn’t start at 1, my brain quietly slides off the page and I start counting wrong. The good news is there’s a small handful of rules that catch almost every off-by-one.
First, your arithmetic example
– Sum from k=1 to 4 of (2k+1): first term is 3, last is 9, number of terms is 4. Sum = 4*(3+9)/2 = 24. All good.
– Sum from k=0 to 4 of (2k+1): first is 1, last is 9.
The key thing you spotted: the number of terms is not 4; it’s 5. From 0 to 4 inclusive there are 5 integers.
Sum = 5*(1+9)/2 = 25.
So the only mistake there was mixing up “n” the upper index with “n” the number of terms. I try to reserve a different letter (say N) for “number of terms,” just to protect myself.
How to get the number of terms every time
– If you’re summing over consecutive integers k = L, L+1, …, U, then
N = U − L + 1.
That’s it. It’s the same whether L starts at 0, 1, or 57.
– Then:
first term = f(L)
last term = f(U)
For an arithmetic progression (constant step in the terms), sum = N*(first + last)/2.
A quick check with your examples:
– k = 1 to 4: N = 4 − 1 + 1 = 4
– k = 0 to 4: N = 4 − 0 + 1 = 5
Now, the geometric sum and reindexing
You had S = sum_{k=2}^{n} 3*(1/2)^k. Your reindexing was exactly right:
– Let j = k − 2. Then when k = 2, j = 0; when k = n, j = n − 2.
– The summand becomes 3*(1/2)^{j+2} = (3/4)*(1/2)^j.
– So S = (3/4) * sum_{j=0}^{n−2} (1/2)^j.
Now use the geometric series formula starting at j = 0:
sum_{j=0}^{M} r^j = (1 − r^{M+1}) / (1 − r).
Here r = 1/2 and M = n − 2, so
sum_{j=0}^{n−2} (1/2)^j = (1 − (1/2)^{(n−2)+1}) / (1 − 1/2)
= (1 − (1/2)^{n−1}) / (1/2).
Multiply the prefactor (3/4):
S = (3/4) * [(1 − (1/2)^{n−1}) / (1/2)]
= (3/2) * [1 − (1/2)^{n−1}].
So your n−1 exponent is correct. If you like a sanity check, plug n = 2:
left side: 3*(1/2)^2 = 3/4
right side: (3/2)*(1 − (1/2)^{1}) = (3/2)*(1/2) = 3/4.
A second way to avoid reindexing at all
There’s a very handy version of the geometric formula you can use directly:
For r ≠ 1,
sum_{k=m}^{n} a*r^k = a*r^m * (1 − r^{n − m + 1}) / (1 − r).
In your problem, a = 3, r = 1/2, m = 2, so
S = 3*(1/2)^2 * (1 − (1/2)^{n − 2 + 1}) / (1 − 1/2)
= (3/4) * (1 − (1/2)^{n−1}) / (1/2)
= (3/2) * (1 − (1/2)^{n−1}).
Same result, no reindexing needed.
A small mental checklist to avoid off-by-one gremlins
– Separate symbols: let N mean “number of terms.” If the upper limit of the sum is also called n, don’t reuse n to mean “how many terms.”
– Count terms right away: N = U − L + 1 for k = L to U.
– Identify first and last terms as f(L) and f(U). Don’t guess from the pattern; plug in the endpoints.
– Arithmetic sums: if the terms step by a constant amount, use N*(first + last)/2.
– Geometric sums:
• If you write it as sum_{j=0}^{M} r^j, remember the formula has r^{M+1}.
• Or use the direct formula from k = m to n: a*r^m * (1 − r^{n − m + 1}) / (1 − r).
– Sanity checks:
• Plug a small upper bound, like n = 2 or 3, and compare with the expanded sum.
• If |r| < 1, the finite sum should approach a finite limit as n grows. For your series, as n → ∞ it tends to 3/2, which matches the formula’s limit. Where your geometric attempt went fuzzy It didn’t, really. The only wobble was second-guessing the exponent. The exponent n−1 came from M+1 with M = n−2 after reindexing, which is exactly right. And counting terms directly also says there are N = n − 1 terms from k = 2 to k = n, which matches the exponent on r in the closed form. I still pause and write N = U − L + 1 in the margin almost every time. It’s boring, but it saves me from the Row 0 problem.
Count terms, don’t guess: in sum from k=a to b, the number of terms is b−a+1 (so for k=0..4 it’s 5, not 4). For reindexing, push the endpoints through the change (j=k−2 gives j=0..n−2) and use sum_{j=0}^{M} r^j = (1−r^{M+1})/(1−r), which here yields sum_{k=2}^{n} 3(1/2)^k = (3/2)(1−(1/2)^{n−1}); your only slip was miscounting terms-the n−1 is correct.