Have these “calculus-based” consecutive summations been discovered yet?












4














Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdotint_0^{n}y_1(x),dx cdot y_1^prime(n) tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdotint_0^{n}y_2(x),dx cdot frac{y_2^prime(n)+1}{y_2^prime(n)} tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$



The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?



Credits for the condensation and formatting of my answer goes to Blue (in the comments)






calculus sequences-and-series







share|cite|improve this question









New contributor




Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
    – Martin R
    15 hours ago












  • @MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
    – Jainam Shah
    15 hours ago


















4














Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdotint_0^{n}y_1(x),dx cdot y_1^prime(n) tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdotint_0^{n}y_2(x),dx cdot frac{y_2^prime(n)+1}{y_2^prime(n)} tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$



The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?



Credits for the condensation and formatting of my answer goes to Blue (in the comments)






calculus sequences-and-series







share|cite|improve this question









New contributor




Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 2




    The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
    – Martin R
    15 hours ago












  • @MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
    – Jainam Shah
    15 hours ago
















4












4








4


2





Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdotint_0^{n}y_1(x),dx cdot y_1^prime(n) tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdotint_0^{n}y_2(x),dx cdot frac{y_2^prime(n)+1}{y_2^prime(n)} tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$



The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?



Credits for the condensation and formatting of my answer goes to Blue (in the comments)






calculus sequences-and-series







share|cite|improve this question









New contributor




Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdotint_0^{n}y_1(x),dx cdot y_1^prime(n) tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdotint_0^{n}y_2(x),dx cdot frac{y_2^prime(n)+1}{y_2^prime(n)} tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$



The only thing is that I'm not quite sure why these relations can include derivatives and integrals and somehow even work at all - they were simply a result of an extremely tiresome trial and error process. And can this "template" be extended to any $S_i(n)$ series or even more unorthodox variants?



Credits for the condensation and formatting of my answer goes to Blue (in the comments)






calculus sequences-and-series




calculus sequences-and-series






share|cite|improve this question









New contributor




Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 14 hours ago





















New contributor




Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 16 hours ago









Jainam Shah

213




213




New contributor




Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Jainam Shah is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 2




    The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
    – Martin R
    15 hours ago












  • @MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
    – Jainam Shah
    15 hours ago
















  • 2




    The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
    – Martin R
    15 hours ago












  • @MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
    – Jainam Shah
    15 hours ago










2




2




The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
– Martin R
15 hours ago






The Euler–Maclaurin formula might be of interest in this context, it describes the connection between the sum $sum_{i=0}^n f(i)$ and the integral $int_0^n f(x) dx$
– Martin R
15 hours ago














@MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
– Jainam Shah
15 hours ago






@MartinR I just had a quick look and I think I might need to spend more time decrypting the logic, since the terminology is much more complex than I can possibly understand. But thanks for pointing me in the right direction!
– Jainam Shah
15 hours ago












1 Answer
1






active

oldest

votes


















3














(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.




Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]




Does this capture the intent of the question?



(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)






share|cite|improve this answer























  • Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
    – Jainam Shah
    15 hours ago












  • Sure. I just noticed you added the case for cubes. Give me a minute.
    – Blue
    15 hours ago










  • You are amazing - thanks a lot!
    – Jainam Shah
    15 hours ago






  • 3




    @JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
    – Blue
    15 hours ago








  • 2




    Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
    – Jainam Shah
    14 hours ago











Your Answer





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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.




Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]




Does this capture the intent of the question?



(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)






share|cite|improve this answer























  • Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
    – Jainam Shah
    15 hours ago












  • Sure. I just noticed you added the case for cubes. Give me a minute.
    – Blue
    15 hours ago










  • You are amazing - thanks a lot!
    – Jainam Shah
    15 hours ago






  • 3




    @JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
    – Blue
    15 hours ago








  • 2




    Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
    – Jainam Shah
    14 hours ago
















3














(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.




Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]




Does this capture the intent of the question?



(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)






share|cite|improve this answer























  • Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
    – Jainam Shah
    15 hours ago












  • Sure. I just noticed you added the case for cubes. Give me a minute.
    – Blue
    15 hours ago










  • You are amazing - thanks a lot!
    – Jainam Shah
    15 hours ago






  • 3




    @JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
    – Blue
    15 hours ago








  • 2




    Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
    – Jainam Shah
    14 hours ago














3












3








3






(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.




Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]




Does this capture the intent of the question?



(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)






share|cite|improve this answer














(Too long for a comment.) The reuse and ambiguity of symbols is a little confusing. I'd like to try to restate the problem.




Consider the sum-of-powers series



$$S_i(n) = 1^i + 2^i + cdots + n^i = sum_{k=1}^n y_i(k) quadtext{with}; y_i(x)=x^i tag{1}$$



For $i=1, 2, 3$, it turns out that we can write
$$begin{align}
S_1(n) &= frac{n+1}{n}cdot y_1^prime(n)cdotint_0^{n}y_1(x),dx tag{2a}\[4pt]
S_2(n) &= frac{n+1}{n}cdot frac{y_2^prime(n)+1}{y_2^prime(n)}cdotint_0^{n}y_2(x),dx tag{2b}\[4pt]
S_3(n) &= left(frac{n+1}{n}right)^2cdotint_0^{n}y_3(x),dx tag{2c}\[4pt]
end{align}$$
which simplify to the well-known formulas
$$begin{align}
S_1(n) &= frac12 n(n+1) tag{3a}\[4pt]
S_2(n) &= frac16 n(n+1)(2n+1) tag{3b}\[4pt]
S_3(n) &= frac14 n^2 (n + 1)^2 tag{3c}
end{align}$$
[Can this technique be extended? Is it new? etc, etc, etc]




Does this capture the intent of the question?



(BTW: I think you need to be a bit more explicit about why the relations in $(2)$ hold.)







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 15 hours ago

























answered 15 hours ago









Blue

47.6k870151




47.6k870151












  • Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
    – Jainam Shah
    15 hours ago












  • Sure. I just noticed you added the case for cubes. Give me a minute.
    – Blue
    15 hours ago










  • You are amazing - thanks a lot!
    – Jainam Shah
    15 hours ago






  • 3




    @JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
    – Blue
    15 hours ago








  • 2




    Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
    – Jainam Shah
    14 hours ago


















  • Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
    – Jainam Shah
    15 hours ago












  • Sure. I just noticed you added the case for cubes. Give me a minute.
    – Blue
    15 hours ago










  • You are amazing - thanks a lot!
    – Jainam Shah
    15 hours ago






  • 3




    @JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
    – Blue
    15 hours ago








  • 2




    Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
    – Jainam Shah
    14 hours ago
















Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
– Jainam Shah
15 hours ago






Thank you Blue for restating my question much more simply - I'm new on this platform so am still learning how to use the syntax. Would you mind completing your edit for the bit that I've added, and I'll replace my text with yours. Is that okay? Thanks again!
– Jainam Shah
15 hours ago














Sure. I just noticed you added the case for cubes. Give me a minute.
– Blue
15 hours ago




Sure. I just noticed you added the case for cubes. Give me a minute.
– Blue
15 hours ago












You are amazing - thanks a lot!
– Jainam Shah
15 hours ago




You are amazing - thanks a lot!
– Jainam Shah
15 hours ago




3




3




@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
– Blue
15 hours ago






@JainamShah: Your question is now clear on where you need help. As Martin suggests, Euler seems to have beaten you to this kind of analysis. (Euler has probably beaten most mathematicians to something.)
– Blue
15 hours ago






2




2




Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
– Jainam Shah
14 hours ago




Gotta give it to him - Euler has not left a single stone unturned. Ah, I tried.
– Jainam Shah
14 hours ago










Jainam Shah is a new contributor. Be nice, and check out our Code of Conduct.










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Jainam Shah is a new contributor. Be nice, and check out our Code of Conduct.













Jainam Shah is a new contributor. Be nice, and check out our Code of Conduct.












Jainam Shah is a new contributor. Be nice, and check out our Code of Conduct.
















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