Uploaded August 2026 | Updated September 2026, 3 weeks ago
In this video, I approximate the accuracy of the partial sums of a series that converges via the Integral Test by estimating the size of the remainder. This is done by comparing the areas of the rectangles below the curve whose function corresponds to the terms of the series, but doing so for all the terms after the n-th partial sum. The areas of the rectangles will be greater than the improper integral starting from n and less than the improper integral starting from n + 1; this is the Remainder Estimate for the Integral Test!
I illustrate this with an example on the series with terms 1/n³ and show that with 32 terms in the partial sum, the accuracy is within 0.0005 of the actual sum of the infinite series.
#math #calculus #series #remainder #education
Timestamps:
- Estimating the sum of a series – 0:00
- Estimating the size of the remainder of a partial sum – 0:35
- Illustrating terms remaining as rectangles below and above a curve – 1:52
- Theorem 2: Remainder estimate for the integral test – 6:19
- Example 5: Approximate the sum of the series with terms 1/n³ – 8:30
- Solution: Obtain the improper integral which equals 1/2n² – 8:55
- Part (a): Partial sum of 10 terms is equal to 1.1975 – 13:10
- By Theorem 2, the remainder is less than 1/200 = 0.005 – 14:54
- Part (b): Want approximation to be accurate within 0.0005 – 16:15
- Solving inequality via Theorem 2 to obtain the minimum number of terms required – 17:39
- We need 32 terms in our partial sum approximation to be within 0.0005 accuracy – 19:00
Notes and playlists:
- Hive: 3speak.tv/watch?v=mes/remainder-estimate-for-the-261
- Notes: peakd.com/mathematics/@mes/infinite-sequences-and-series-the-integral-test-and-estimate-of-sums
- Playlist: youtube.com/playlist?list=PLTXCLnbwZspA
- Sequences and Series: youtube.com/playlist?list=PLai3U8-WIK0EXHAJ3vRg0T_kKEyPah1Lz
- MES Links: https://mes.fm/links .
------------------------------------------------------
MES Links: https://mes.fm/links
Donate: https://mes.fm/donate
MES Truth: youtube.com/@mestruth
Official Website: https://MES.fm
Hive: peakd.com/@mes
Email me: contact@mes.fm
Percentage Calculator: https://percentagecalculator.mes.fm
Grade Calculator: https://gradecalculator.mes.fm
BMI Calculator: https://bmicalculator.mes.fm
Mortgage Calculator: https://mortgagecalculator.mes.fm
Timer: https://timer.mes.fm/
In this video, I approximate the accuracy of the partial sums of a series that converges via the Integral Test by estimating the size of the remainder. This is done by comparing the areas of the rectangles below the curve whose function corresponds to the terms of the series, but doing so for all the terms after the n-th partial sum. The areas of the rectangles will be greater than the improper integral starting from n and less than the improper integral starting from n + 1; this is the Remainder Estimate for the Integral Test!
I illustrate this with an example on the series with terms 1/n³ and show that with 32 terms in the partial sum, the accuracy is within 0.0005 of the actual sum of the infinite series.
#math #calculus #series #remainder #education
Timestamps:
- Estimating the sum of a series – 0:00
- Estimating the size of the remainder of a partial sum – 0:35
- Illustrating terms remaining as rectangles below and above a curve – 1:52
- Theorem 2: Remainder estimate for the integral test – 6:19
- Example 5: Approximate the sum of the series with terms 1/n³ – 8:30
- Solution: Obtain the improper integral which equals 1/2n² – 8:55
- Part (a): Partial sum of 10 terms is equal to 1.1975 – 13:10
- By Theorem 2, the remainder is less than 1/200 = 0.005 – 14:54
- Part (b): Want approximation to be accurate within 0.0005 – 16:15
- Solving inequality via Theorem 2 to obtain the minimum number of terms required – 17:39
- We need 32 terms in our partial sum approximation to be within 0.0005 accuracy – 19:00
Notes and playlists:
- Hive: 3speak.tv/watch?v=mes/remainder-estimate-for-the-261
- Notes: peakd.com/mathematics/@mes/infinite-sequences-and-series-the-integral-test-and-estimate-of-sums
- Playlist: youtube.com/playlist?list=PLTXCLnbwZspA
- Sequences and Series: youtube.com/playlist?list=PLai3U8-WIK0EXHAJ3vRg0T_kKEyPah1Lz
- MES Links: https://mes.fm/links .
------------------------------------------------------
MES Links: https://mes.fm/links
Donate: https://mes.fm/donate
MES Truth: youtube.com/@mestruth
Official Website: https://MES.fm
Hive: peakd.com/@mes
Email me: contact@mes.fm
Percentage Calculator: https://percentagecalculator.mes.fm
Grade Calculator: https://gradecalculator.mes.fm
BMI Calculator: https://bmicalculator.mes.fm
Mortgage Calculator: https://mortgagecalculator.mes.fm
Timer: https://timer.mes.fm/

![Vortex Math is Based on the Modulo Multiplication Identity
In this video I show that the previous result of multiplying a number and its vortex sum yields the same vortex sum can be written in its equivalent modulo multiplication identity. Since we have already established that the vortex sum of an integer (summing the digits until we get a single digit) is the same as the modulo of that number with the modulus being the base - 1, we can rewrite the vortex sums using the modulo operations. In general, if we have integers A, B, and C, then we have the identity AB mod C = [A·(B mod C)] mod C. Furthermore we can apply this same identity with the integers inside the bracket to also get it equal to [( A mod C)(B mod C)] mod C and [(A mod C)·B] mod C. Pretty epic stuff!
#math #vortexmath #modulararithmetic #numbertheory #education
Timestamps:
- Vortex math multiplication as an equivalent modulo identity – 0:00
- 25 * 4 = 100 v= 1 or 25 v= 7 * 4 = 28 v= 1 – 0:38
- (4*25) mod 9 = [4*(25 mod 9)] mod 9 = 1 – 1:59
- Generalize modulo operation to any integer multiple or modulus: (AB mod C = [A(B mod C)] mod C = [(A mod C)(B mod C)] mod C = [(A mod C)B] mod C – 3:00
- Example: A = 11, B = 22, C = 7 – 4:44
- Reason is because modulo operation gives integers, which can just apply the same identity over again – 5:51
– Modulo of a multiplication of integers is the same as the modulo of the multiplication of either or both modulo of the integers – 6:54
Notes and playlists:
- Summary: https://inleo.io/threads/view/mes/re-leothreads-2epxg3553
- Notes: https://peakd.com/hive-128780/@mes/messcience-2-vortex-math-part-1-number-theory-and-modular-arithmetic
- Vortex Math playlist: https://www.youtube.com/playlist?list=PLai3U8-WIK0EbRnMsUBx2RxlerL7GQuLX
- MES Science playlist: https://www.youtube.com/playlist?list=PLai3U8-WIK0GhjCHmTw1XbqMD_EdVKdd9 .
Become a MES Super Fan! https://www.youtube.com/channel/UCUUBq1GPBvvGNz7dpgO14Ow/join
DONATE! ʕ •ᴥ•ʔ https://mes.fm/donate
SUBSCRIBE via EMAIL: https://mes.fm/subscribe
MES Links: https://mes.fm/links
MES Truth: https://mes.fm/truth
Official Website: https://MES.fm
Hive: https://peakd.com/@mes
Email me: contact@mes.fm
Free Calculators: https://mes.fm/calculators
BMI Calculator: https://bmicalculator.mes.fm
Grade Calculator: https://gradecalculator.mes.fm
Mortgage Calculator: https://mortgagecalculator.mes.fm
Percentage Calculator: https://percentagecalculator.mes.fm
Free Online Tools: https://mes.fm/tools
iPhone and Android Apps: https://mes.fm/mobile-apps Vortex Math is Based on the Modulo Multiplication Identity](https://i.ytimg.com/vi/cJTiB6njXAc/mqdefault.jpg)








