# Can we prove summation formula for the first $n$ terms of natural numbers through calculus? [closed]

Can we prove summation formula for the first $n$ terms of natural numbers through calculus?

What about the summation of first $n$ numbers of the form $a^k$ and other summation formulas like sum of a GP or AGP?

Essentially using calculus please.

## closed as unclear what you're asking by Jack, Rob Arthan, Claude Leibovici, user91500, WatsonOct 16 '16 at 8:46

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• You need difference equations, analogous to—but different from—derivatives. – Ted Shifrin Oct 16 '16 at 0:53
• See Faulhaber's formula. You can search this site for it as well. – Ross Millikan Oct 16 '16 at 0:55
• Do you mean $1^k+2^k+\dots+n^k$ or do you mean $a^1+a^2+\dots+a^n$? – Simply Beautiful Art Oct 16 '16 at 1:09
• @SimpleArt both – ankit Oct 16 '16 at 1:11
• For $1^k+2^k+\dots+n^k$. – Simply Beautiful Art Oct 16 '16 at 1:27

For the sort of arithmetic progression:

If you consider some function $f_k(x)$ that satisfies

1. $f_k(x)=f(x-1)+x^k$

2. $f_k(1)=1$

Then for all $x\in\mathbb N$, $f(x)=\sum_{n=1}^xn^k$.

$$f_k(x)=f_k(x-1)+x^k$$

$$f'_k(x)=f'_k(x-1)+kx^{k-1}$$

$$f'_k(x)=f'_k(x-2)+k\left[(x-1)^{k-1}+x^{k-1}\right]\\\vdots\\ f'_k(x)=f'_k(0)+k\sum_{n=1}^xn^{k-1}\tag{x\in\mathbb N}$$

$$f'_k(x)=f'_k(0)+kf_{k-1}(x)$$

\begin{align}f_k(x)-\require{cancel}\cancelto0{f_k(0)}&=\int_0^xf'_k(t)dt\\&=\int_0^xf'_k(0)+kf_{k-1}(t)dt\\&=f'(0)x+k\int_0^xf_{k-1}(t)dt\end{align}

which sets a recursive formula for finding $f_k(x)$. Since it is relatively clear that $f_0(x)=x$, one can derive $f_k(x)$ for all $k\in\mathbb N$. To find $f'_k(0)$, plug in $x=1$.

$$\sum_{n=0}^xr^n=\frac{1-r^{x+1}}{1-r}$$

$$f_0(x)=\lim_{r\to1}\frac{1-r^{x+1}}{1-r}$$

$$\frac{d}{dr}\sum_{n=0}^xr^n=\sum_{n=0}^xnr^{n-1}=\frac{d}{dr}\frac{1-r^{x+1}}{1-r}\tag{AGP?}$$

$$f_1(x)=\lim_{r\to1}\frac{d}{dr}\frac{1-r^{x+1}}{1-r}$$

In general,

$$f_k(x)=\lim_{r\to1}\underbrace{\left(\frac{d}{dr}r\left(\frac{d}{dr}r\left(\dots r\left(\frac{d}{dr}\frac{1-r^{x+1}}{1-r}\right)\dots\right)\right)\right)}_{k\ \frac{d}{dr}'s}$$

For the geometric progression:

$$\frac{1-r^{x+1}}{1-r}=\frac1{1-r}-\frac{r^{x+1}}{1-r}$$

Take the taylor expansion of both expressions around $r=0$ and you will get

$$=\left(1+r+r^2+\dots+r^x+\color{red}{r^{x+1}+\dots}\right)-\left(\color{red}{r^{x+1}+r^{x+2}+\dots}\right)$$

$$=1+r+r^2+\dots+r^x$$

Doing this backwards would result in solving a differential equation at the step where we take the taylor expansion.

Well if you mean $a^1+a^2+...+a^n$, just plug in the arithmetic progression formula. If you want to calculate $1^k+2^k+...+n^k$, here is an iterative method that you may apply. For instance, when we are calculating $n^3$, in fact $(n+3)(n+2)(n+1)n-(n+2)(n+1)n(n-1)=4(n+2)(n+1)n=4(n^3+3n^2+2n)$ $$\sum_{i=1}^n (n+3)(n+2)(n+1)n-(n+2)(n+1)n(n-1)=\sum_{i=1}^n 4(n^3+3n^2+2n)$$ We can subtract the sum of $n^2$ and $n$ part to get the sum of $n^3$ part.

• Um, you mean geometric progression in the first line? And the OP seems to know this and wants this done with calculus. – Simply Beautiful Art Oct 16 '16 at 1:28
• Yes as Simple Art says I know it. that is why I tagged it with "alternative-proof" – ankit Oct 16 '16 at 2:13