prove ramanujan summation series upto infinity
Answers
Answer:
For those of you who are unfamiliar with this series, which has come to be known as the Ramanujan Summation after a famous Indian mathematician named Srinivasa Ramanujan, it states that if you add all the natural numbers, that is 1, 2, 3, 4, and so on, all the way to infinity, you will find that it is equal to -1/12.
Step-by-step explanation:
Ramanujan summation is a technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series. Although the Ramanujan summation of a divergent series is not a sum in the traditional sense, it has properties which make it mathematically useful in the study of divergent infinite series, for which conventional summation is undefined.
Summation Edit
Ramanujan summation essentially is a property of the partial sums, rather than a property of the entire sum, as that doesn't exist. If we take the Euler–Maclaurin summation formula together with the correction rule using Bernoulli numbers, we see that:
{\displaystyle {\begin{aligned}{\frac {1}{2}}f(0)+f(1)+\cdots +f(n-1)+{\frac {1}{2}}f(n)&={\frac {1}{2}}[f(0)+f(n)]+\sum _{k=1}^{n-1}f(k)\\&=\int _{0}^{n}f(x)\,dx+\sum _{k=1}^{p}{\frac {B_{k+1}}{(k+1)!}}\left[f^{(k)}(n)-f^{(k)}(0)\right]+R_{p}\end{aligned}}}{\displaystyle {\begin{aligned}{\frac {1}{2}}f(0)+f(1)+\cdots +f(n-1)+{\frac {1}{2}}f(n)&={\frac {1}{2}}[f(0)+f(n)]+\sum _{k=1}^{n-1}f(k)\\&=\int _{0}^{n}f(x)\,dx+\sum _{k=1}^{p}{\frac {B_{k+1}}{(k+1)!}}\left[f^{(k)}(n)-f^{(k)}(0)\right]+R_{p}\end{aligned}}}
Ramanujan[1] wrote it for the case p going to infinity:
{\displaystyle \sum _{k=1}^{x}f(k)=C+\int _{0}^{x}f(t)\,dt+{\frac {1}{2}}f(x)+\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(x)}{\displaystyle \sum _{k=1}^{x}f(k)=C+\int _{0}^{x}f(t)\,dt+{\frac {1}{2}}f(x)+\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(x)}
where C is a constant specific to the series and its analytic continuation and the limits on the integral were not specified by Ramanujan, but presumably they were as given above. Comparing both formulae and assuming that R tends to 0 as x tends to infinity, we see that, in a general case, for functions f(x) with no divergence at x = 0:
{\displaystyle C(a)=\int _{0}^{a}f(t)\,dt-{\frac {1}{2}}f(0)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(0)}{\displaystyle C(a)=\int _{0}^{a}f(t)\,dt-{\frac {1}{2}}f(0)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(0)}
where Ramanujan assumed {\displaystyle a=0.}{\displaystyle a=0.} By taking {\displaystyle a=\infty }{\displaystyle a=\infty } we normally recover the usual summation for convergent series. For functions f(x) with no divergence at x = 1, we obtain:
{\displaystyle C(a)=\int _{1}^{a}f(t)\,dt+{\frac {1}{2}}f(1)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(1)}{\displaystyle C(a)=\int _{1}^{a}f(t)\,dt+{\frac {1}{2}}f(1)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(1)}
C(0) was then proposed to use as the sum of the divergent sequence. It is like a bridge between summation and integration.
The convergent version of summation for functions with appropriate growth condition is then:
{\displaystyle f(1)+f(2)+f(3)+\cdots =-{\frac {f(0)}{2}}+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{e^{2\pi t}-1}}\,dt}{\displaystyle f(1)+f(2)+f(3)+\cdots =-{\frac {f(0)}{2}}+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{e^{2\pi t}-1}}\,dt}
To compare, see Abel–Plana formula.